chap15

PART 2 OSCILLATIONS, WAVES, AND FLUIDSCHAPTER 15 OSCILLATORY MOTION ActivPhysics can help with these problems: All Activities in Section 9 Sections 15-1 and 15-2: Oscillations and Simple Harmonic Motion Problem 1. A doctor counts 77 heartbeats in one minute. What are the period and frequency of the heart’s oscillations? Solution If 77 heartbeats take 1 min., then one heartbeat (one cycle) takes T = 1 min/77 = 60 s/77 = 0.77 s, which is the period. The frequency is f = 77/ min = 77/60 s = 1.28 Hz. (One can see that T = 1/f.) Problem 2. A violin string playing the note “A” oscillates at 440 Hz. What is the period of its oscillation? Solution T = 1/f = 1/440 Hz = 2.27 ms (Equation 15-1). Problem 3. The vibration frequency of a hydrogen chloride molecule is 8.66×10 13 Hz. How long does it take the molecule to complete one oscillation? Solution T = 1/f = 1/(8.66×10 13 Hz) = 1.15×10 −14 s = 11.5 fs (Equation 15-1). Problem 4. Write expressions for simple harmonic motion (a) with amplitude 10 cm, frequency 5.0 Hz, and with maximum displacement at t = 0 and (b) with amplitude 2.5 cm, angular frequency 5.0 s −1 , and with maximum velocity at t = 0. Solution (a) Use Equation 15-9 with A = 10 cm, ω = 2π(5 Hz) = 10π s −1 and φ = 0. Then x(t) = (10 cm)× cos[(10π s −1 )t]. (b) The maximum (positive) velocity occurs at t = 0 if sin φ = −1 (from Equation 15-10), therefore, the motion is described by Equation 15-9 with A = 2.5 cm, ω = 5 s −1 , and φ = −π/2. Since cos(ωt −π/2) = sin ωt, the expression simplifies to x(t) = (2.5 cm) sin[(5 s −1 )t]. Problem 5. Determine the amplitude, angular frequency, and phase constant for each of the simple harmonic motions shown in Fig. 15-33. Solution The amplitude is the maximum displacement, read along the x axis (ordinate) in Fig. 15-33. The angular frequency is 2π times the reciprocal of the period, which is the time interval between corresponding points read along the t axis (abscissa). The phase constant can be determined from the intercept and slope (displacement and velocity) at t = 0. One sees that (a) A ≃ 20 cm, ω ≃ 2π/4 s ≃ 1 2 π s −1 , and φ ≃ 0; (b) A ≃ 30 cm, ω ≃ 2π/3.2 s ≃ 2 s −1 , and φ ≃ −90 ◦ ≃ − 1 2 π; (c) A ≃ 40 cm, ω ≃ 2π/(2×2 s) ≃ 1 2 π s −1 , and φ ≃ cos −1 (27/40) ≃ 48 ◦ ≃ 1 4 π. figure 15-33 Problem 5 Solution. Problem 6. A 200-g mass is attached to a spring of constant k = 5.6 N/m and set into oscillation with amplitude A = 25 cm. Determine (a) the frequency in hertz, (b) the period, (c) the maximum velocity, and (d) the maximum force in the spring. 232 CHAPTER 15 Solution For a mass on a spring, Equation 15-8a gives ω = k/m = (5.6 N/m)/(0.2 kg) = 5.29 s −1 . Then (a) f = ω/2π = 0.842 Hz (Equation 15-8b), (b) T = 1/f = 1.19 s (Equation 15-8c), (c) v max = ωA = (5.29 s −1 )(0.25 m) = 1.32 m/s (Equation 15-10), and (d) F max = ma max = mω 2 A = kx max = kA = (5.6 N/m)(0.25 m) = 1.40 N (Equations 15-2 or 3 or 11). Problem 7. An astronaut in an orbiting spacecraft is “weighed” by being strapped to a spring of constant k = 400 N/m, and set into simple harmonic motion. If the oscillation period is 2.5 s, what is the astronaut’s mass? Solution We suppose that the other end of the spring is fastened to the spacecraft, whose mass is much greater than the astronaut’s, and that the orbiting system is approximately inertial. Then Equation 15-8c gives m = k/ω 2 = k(T/2π) 2 = (400 N/m)(2.5 s/2π) 2 = 63.3 kg. Problem 8. A simple model for an automobile suspension consists of a mass attached to a spring. If the mass is 1900 kg and the spring constant is 26 kN/m, with what frequency and period will the car undergo simple harmonic motion? Solution Equations 15-8b and 8c give f = (1/2π) k/m = 0.589 Hz, and T = 1/f = 1.70 s. Problem 9. A simple model of a carbon dioxide (CO 2 ) molecule consists of three mass points (the atoms) connected by two springs (electrical forces), as suggested in Fig. 15-34. One way this system can oscillate is if the carbon atom stays fixed and the two oxygens move symmetrically on either side of it. If the frequency of this oscillation is 4.0×10 13 Hz, what is the effective spring constant? The mass of an oxygen atom is 16 u. figure 15-34 Problem 9. Solution With the carbon atom end of either “spring” fixed, the frequency of either oxygen atom is ω = 2πf = k/m. Therefore k =(2π×4×10 13 Hz) 2 (16×1.66×10 −27 kg) = 1.68×10 3 N/m. Problem 10. Sketch the following simple harmonic motions on the same graph: (a) x = (15 cm)[cos(2.5t +π/2)], (b) motion with amplitude 30 cm, period 5.0 s, phase constant 0; (c) motion with amplitude 15 cm, frequency 0.40 Hz, phase constant 0. Solution (a) This equation has amplitude 15 cm, period T = 2π/(2.5 s −1 ) = 2.51 s (t in seconds is understood), and phase constant 1 2 π. Note that cos(ωt + 1 2 π) = −sin ωt. (b) and (c) both are represented by equations of the form x = Acos ωt, but the amplitude and period of (c) are half those of (b). Except for the difference in phase, (c) is nearly the same as (a). Problem 10 Solution. Problem 11. Two identical mass-spring systems consist of 430-g masses on springs of constant k = 2.2 N/m. Both are displaced from equilibrium and the first released at time t = 0. How much later should the second be released so the two oscillations differ in phase by π/2? Solution Suppose that both masses are released from their maximum positive displacements, the first at t = 0 and the second at t = t 0 . Then, since the phase (the argument of the cosine in Equation 15-9) of each motion is zero at release, φ 1 = 0 and ω 2 t 0 +φ 2 = 0. The difference in phase is 1 2 π = (ω 1 t +φ 1 ) −(ω 2 t + φ 2 ) = (ω 1 −ω 2 )t + (φ 1 −φ 2 ) = −φ 2 , where ω 1 = ω 2 = k/m (for identical mass-spring systems). Thus, t 0 = −φ 2 /ω 2 = 1 2 π m/k = 1 2 π (0.43 kg)/(2.2 N/m) = 0.694 s. (In terms of the period, t 0 = −φ 2 /(2π/T) = 1 4 T, which is intuitively more obvious.) CHAPTER 15 233 Problem 12. The quartz crystal in a digital quartz watch executes simple harmonic motion at 32,768 Hz. (This is 2 15 Hz, chosen so 15 divisions by 2 give a signal at 1.00000 Hz.) If each face of the crystal undergoes a maximum displacement of 100 nm, find the maximum velocity and acceleration of the crystal faces. Solution The amplitude is A = 10 −7 m, and the angular frequency is ω = 2π(32,768 Hz) = 2.06×10 5 s −1 . Therefore, v max = Aω = 2.06 cm/s, and a max = Aω 2 = ωv max = 4.24 km/s 2 . Problem 13. A mass m slides along a frictionless horizontal surface at speed v 0 . It strikes a spring of constant k attached to a rigid wall, as shown in Fig. 15-35. After a completely elastic encounter with the spring, the mass heads back in the direction it came from. In terms of k, m, and v 0 , determine (a) how long the mass is in contact with the spring and (b) the maximum compression of the spring. m v 0 figure 15-35 Problem 13. Solution (a) While the mass is in contact with the spring, the net horizontal force on it is just the spring force, so it undergoes half a cycle of simple harmonic motion before leaving the spring with speed v 0 to the left. This takes time equal to half a period 1 2 T = π/ω = π m/k. (b) v 0 is the maximum speed, which is related to the maximum compression of the spring (the amplitude) by v 0 = ωA. Thus A = v 0 /ω = v 0 m/k. Problem 14. A 50-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 15 m/s 2 and its maximum speed is 3.5 m/s. Determine (a) the angular frequency, (b) the spring constant, and (c) the amplitude of the motion. Solution (a) From a max = Aω 2 and v max = Aω, we find ω = a max /v max = (15 m/s 2 )/(3.5 m/s) = 4.29 s −1 . (b) k = mω 2 = (0.05 kg)(4.29 s −1 ) 2 = 0.918 N/m. (c) A = v max /ω = a max /ω 2 = v 2 max /a max = (etc.) = (3.5 m/s) 2 /(15 m/s 2 ) = 0.817 m (the latter expression, in terms of given data, is preferred). Problem 15. Show by substitution that x(t) = Asin ωt is a solution to Equation 15-4. Solution If x = Asinωt, dx/dt = ωAcos ωt, and d 2 x/dt 2 = −ω 2 Asin ωt = −ω 2 x. Substituting into Equation 15-4, we find m(−ω 2 x) = −kx, which is satisfied if ω 2 = k/m. Problem 16. Show by substitution that x(t) = a cos ωt −b sinωt is a solution to Equation 15-4, and that this form is equivalent to Equation 15-9 with A = √ a 2 +b 2 and φ = tan −1 (b/a). Solution It is easy enough to substitute the given expression into Equation 15-4, but it is shown in the paragraph before Equation 15-8a and in the solution to the previous problem that cos ωt and sinωt are solutions, respectively. An important property of linear differential equations is that a linear combination of solutions is also a solution. If we expand Equation 15-9, using a trigonometric identity from Appendix A, we find Acos(ωt +φ) = A(cos φcos ωt − sin φsin ωt). If we define a = Acos φ and b = Asinφ, then A = √ a 2 +b 2 and φ = tan −1 (b/a), and the equivalence of the two forms is evident. Section 15-3: Applications of Simple Harmonic Motion Problem 17. How long should you make a simple pendulum so its period is (a) 200 ms; (b) 5.0 s; (c) 2.0 min? Solution The period and length of a simple pendulum (at the surface of the Earth) are related by Equation 15-18, therefore ℓ = (T/2π) 2 g = (0.248 m/s 2 )T 2 = 9.93 mm, 6.21 m, and 3.57 km respectively, for the three values given. 234 CHAPTER 15 Problem 18. At the heart of a grandfather clock is a simple pendulum 1.45 m long; the clock ticks each time the pendulum reaches its maximum displacement in either direction. What is the time interval between ticks? Solution The clock ticks twice each period of oscillation, so the time between ticks is 1 2 T = π ℓ/g = 1.21 s. Problem 19. A 640-g hollow ball 21 cm in diameter is suspended by a wire and is undergoing torsional oscillations at a frequency of 0.78 Hz. What is the torsional constant of the wire? Solution The rotational inertia of a hollow sphere about an axis through its center is 2 3 MR 2 (see Table 12-2). Equation 15-14 gives κ=ω 2 I = (2π×0.78 Hz) 2 × ( 2 3 ×0.64 kg)( 1 2 ×0.21 m) 2 = 0.113 N·m/rad. Problem 20. A physics student, bored by a lecture on simple harmonic motion, idly picks up his pencil (mass 9.2 g, length 17 cm) by the tip with his frictionless fingers, and allows it to swing back and forth with small amplitude. If it completes 6279 full cycles during the lecture, how long does the lecture last? (Hint: See Example 15-5.) Solution Regarding the pencil as a thin rod suspended from one end, as in Example 15-5, we find the period is T = 2π 2ℓ/3 g. The length of the lecture (if we assume that the student became bored the instant it began) is 6279(2π)× 2(0.17 m)/3(9.8 m/s 2 ) = 4.24×10 3 s = 70.7 min . Problem 21. A pendulum of length ℓ is mounted in a rocket. What is its period if the rocket is (a) at rest on its launch pad; (b) accelerating upward with acceleration a = 1 2 g; (c) accelerating downward with acceleration a = 1 2 g; (d) in free fall? Solution (It may be helpful to think of an elevator instead of a rocket in this problem.) Let a be the acceleration of the pendulum relative to the rocket, and let a 0 be the acceleration of the rocket relative to the ground (assumed to be an inertial system). Then Newton’s second law is ¸ F = m(a +a 0 ), or ¸ F−ma 0 = ma. The rotational analog of this equation is the appropriate generalization of Equation 15-15 for a simple pendulum in an accelerating frame. The “fictitious” torque (about the point of suspension), r ×(−ma 0 ), can be combined with the torque of gravity, r ×mg, if we replace g by |g −a 0 |, while the right-hand side, |r ×ma| = Iα, is the same as Equation 15-15. For small oscillations about the equilibrium position (which depends on a 0 ), the period is T = 2π ℓ/ |g −a 0 |. (a) If a 0 = 0, T = 2π ℓ/g, as before. (b) Take the y axis positive up. Then a 0 = 1 2 gˆ and g = −gˆ, so T = 2π ℓ/(g + 1 2 g) = 2π 2ℓ/3g. (c) If a 0 = − 1 2 gˆ, T = 2π ℓ/(g − 1 2 g) = 2π 2ℓ/g. (d) If a 0 = g, T = ∞ (there is no restoring torque and the pendulum does not oscillate). Problem 22. A 340-g mass is attached to a vertical spring and lowered slowly until it rests at a new equilibrium position, which is 30 cm below the spring’s original equilibrium. The system is then set into simple harmonic motion. What is the period of the motion? Solution At the “new” equilibrium position, mg = kx 1 (see the beginning of Section 15-3) so m/k = x 1 /g = 0.3 m÷ (9.8 m/s 2 ) = (T/2π) 2 . Thus T = 1.10 s. Problem 23. A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is in the original unstretched equilibrium position it had before the mass was attached; the low point is 5.8 cm below this. What is the period of oscillation? Solution At the highest point, there is no spring force (since the spring is unstretched), so the acceleration is just g (downward). This is also the maximum acceleration during the simple harmonic motion (since a max occurs where the displacement is maximum), so a max = g = ω 2 A. The peak-to-peak displacement is 2A = 5.8 cm, thus T = 2π/ω = 2π A/g = 2π 0.029 m/(9.8 m/s 2 ) = 0.342 s. CHAPTER 15 235 Problem 24. Derive the period of a simple pendulum by considering the horizontal displacement x and the force acting on the bob, rather than the angular displacement and torque. Solution The tangential component of Newton’s second law, for a simple pendulum of mass m and length ℓ, is m(d 2 s/dt 2 ) = −mg sin θ. (The radial component guarantees that the motion follows a circular arc.) The horizontal displacement is x = ℓ sinθ. For small displacements, x ≈ ℓθ = s, so the equation of motion is approximately m d 2 s dt 2 ≈ m d 2 x dt 2 = −mg x ℓ , or d 2 x dt 2 = − g ℓ x. This is the equation for simple harmonic motion with period T = 2π/ω = 2π ℓ/g. Problem 24 Solution. Problem 25. A solid disk of radius R is suspended from a spring of linear spring constant k and torsional constant κ, as shown in Fig. 15-36. In terms of k and κ, what value of R will give the same period for the vertical and torsional oscillations of this system? Solution Equating the angular frequencies for vertical and torsional oscillations (Equations 15-8a and 15-14), we find k/M = κ/I = κ/( 1 2 MR 2 ), or R = 2κ/k. Problem 26. A thin steel beam 8.0 m long is suspended from a crane and is undergoing torsional oscillations. Two 75-kg steelworkers leap onto opposite ends of the beam, as shown in Fig. 15-37. If the frequency figure 15-36 Problem 25. of torsional oscillations diminishes by 20%, what is the mass of the beam? figure 15-37 Problem 26. Solution The frequencies before and after the steel workers jump, f 2 = 0.8f 1 , can be related to the rotational inertias of the beam, with and without them, through Equation 15-14, f 2 /f 1 = I 1 /I 2 , since the torsional constant κ is not changed. Before, I 1 = 1 12 Mℓ 2 (Table 12-2), and after, I 2 = I 1 + 2m(ℓ/2) 2 = 1 12 ℓ 2 × (M + 6m), where m = 75 kg is each steelworker’s mass. Therefore I 1 = (f 2 /f 1 ) 2 I 2 = 1 12 Mℓ 2 = 1 12 (0.8) 2 ℓ 2 (M + 6m), or M = 6(75 kg)(0.64)÷ (1 −0.64) = 800 kg. Problem 27. Geologists use an instrument called a gravimeter to measure the local acceleration of gravity. A particular gravimeter uses the period of a 1-m-long pendulum to determine g. If g is to be measured to within 1 mgal(1 gal = 1 cm/s 2 ) and if the period can be measured with arbitrary accuracy, how accurately must the length of the pendulum be known? 236 CHAPTER 15 Solution For a simple pendulum, ℓ = (T/2π) 2 g. If the period is known precisely (∆T ≈ 0), then the fractional errors in ℓ and g are the same, i.e., ∆ℓ/ℓ = ∆g/g = 1 mgal÷ 980 gal ≈ 10 −6 . Problem 28. A pendulum consists of a 320-g solid ball 15.0 cm in diameter, suspended by an essentially massless string 80.0 cm long. Calculate the period of this pendulum, treating it first as a simple pendulum and then as a physical pendulum. How much error is introduced by the simple pendulum approximation? Hint: Remember the parallel-axis theorem. Solution For a simple pendulum, T simp = 2π ℓ/g = 2π 0.875 m/9.8 m/s 2 = 1.877 . . . s. The more exact expression for this type of physical pendulum (called Borda’s pendulum) is T phys = 2π I/mgℓ = 2π 2 5 mR 2 +mℓ 2 /mgℓ = T simp 1 + (2R 2 /5ℓ 2 ) = T simp 1 + 0.4(7.5/87.5) 2 = (1.00147 . . .)T simp = 1.880 . . . s. The fractional error is (T phys −T simp )/T phys = 0.147%. Problem 28 Solution. Problem 29. A thin, uniform hoop of mass M and radius R is suspended from a thin horizontal rod and set oscillating with small amplitude, as shown in Fig. 15-38. Show that the period of the oscillations is 2π 2R/g. Hint: You may find the parallel-axis theorem useful. figure 15-38 Problem 29 Solution. Solution Equation 15-16 gives T = 2π I/Mgℓ = 2π (MR 2 +MR 2 )/MgR = 2π 2R/g, where we used the parallel axis theorem for I, with ℓ = h = R. Problem 30. A solid disk of mass M and radius R is mounted on a horizontal axle, as shown in Fig. 15-39. A spring of spring constant k is connected to the disk at a point 1 2 R above the axle, and in equilibrium runs horizontally to a wall. If the disk is rotated slightly away from equilibrium, what is the angular frequency of the resulting oscillations? Hint: For small θ, sin θ ≃ θ and cos θ ≃ 1. Solution For small (approximately horizontal) displacements of the end of the spring, x≈ 1 2 Rθ. The restoring torque of the spring about the central axle, τ ≈ −( 1 2 R)(kx) ≈ − 1 4 kR 2 θ, equals I(d 2 θ/dt 2 ) for the disk. This is the equation for simple harmonic motion with ω = ( 1 4 kR 2 )/I = ( 1 4 kR 2 )/( 1 2 MR 2 ) = k/2M. figure 15-39 Problem 30 Solution. CHAPTER 15 237 Problem 31. A point mass m is attached to the rim of an otherwise uniform solid disk of mass M and radius R (Fig. 15-40). The disk is rolled slightly away from its equilibrium position and released. It rolls back and forth without slipping. Show that the period of this motion is given by T = 2π 3MR/2mg. Solution An exact solution of this problem requires advanced methods in analytical mechanics. However, there are two simple heuristic arguments showing that the period given is approximately correct, for small displacements from equilibrium, if the point mass is also assumed to be much smaller than the mass of the disk, i.e., m ≪M. Consider the disk when it is displaced by a small angle, θ ≪1, from equilibrium, as shown in the sketch. The unbalanced torque about the point of contact, O (which is instantaneously at rest for rolling without slipping) is τ = Iα = −mgRsin θ ≈ −mgRθ. This is a restoring torque because it is in the opposite direction to the angular displacement. If we neglect m compared to M, the rotational inertia is just that of the disk about O, or I = 1 2 MR 2 +MR 2 = 3 2 MR 2 (recall Equation 12-20), so α ≈ −mgRθ/I = −(2mg/3MR)θ. This is the equation for simple harmonic motion (see Equation 15-13) with period T = 2π/ω = 2π× 3MR/2mg (see Equation 15-14). Alternatively, the potential energy varies like that of just the point mass (since that of the disk is constant when it rolls on a horizontal surface), so U = mgy = mgR(1 −cos θ) ≈ 1 2 mgRθ 2 , for small θ. For m ≪M, the kinetic energy is approximately that of just the disk, rolling without slipping, or K = 1 2 Iω 2 = 1 2 ( 3 2 MR 2 )(dθ/dt) 2 . The total energy is E = K +U ≈ 1 2 (mgR)θ 2 + 1 2 ( 3 2 MR 2 )(dθ/dt) 2 . By analogy with a mass-spring system (where E = 1 2 kx 2 + 1 2 m(dx/dt) 2 ), this represents simple harmonic motion with angular frequency (analogous to k/m) of (mgR)/( 3 2 MR 2 ) = 2mg/3MR (see also Problem 44). 870 figure 15-40 Problem 31 Solution. Problem 32. Repeat the previous problem for the case when the disk does not contact the ground but is mounted on a frictionless horizontal axle through its center. Why is your answer different? Solution The net (restoring) torque about the axle is due only to the point mass, therefore τ = −mgRsin θ = Iα, where I = 1 2 MR 2 +mR 2 . For small angles, sin θ ≈θ, so α = −(mgR/I)θ, which is the equation for simple harmonic motion with period T = 2π I/mgR = 2π (R/g)(1 +M/2m). Alternatively, we could use Equation 15-16 for a physical pendulum, whose CM is suspended a distance ℓ = mR/(m+M) from the axle. Then T = 2π I/(m+M)gℓ , as before. This result is not the same as that of the previous problem, because the forces acting are different (there is an axle, there is no friction, etc.). (Also, in this problem there is no restriction on the relative size of the masses, whereas the answer to Problem 31 holds only for m ≪M.) Problem 32 Solution. Problem 33. A cyclist turns her bicycle upside down to tinker with it. After she gets it upside down, she notices the front wheel executing a slow, small-amplitude back-and-forth rotational motion with a period of 12 s. Considering the wheel to be a thin ring of mass 600 g and radius 30 cm, whose only irregularity is the presence of the tire valve stem, determine the mass of the valve stem. 238 CHAPTER 15 Solution The bicycle wheel may be regarded as a physical pendulum, with rotational inertia I = MR 2 +mR 2 about its central axle, where M = 600 g is the mass of the wheel (thin ring) and m is the mass of the valve stem (a circumferential point mass). The distance of the CM from the axle is given by (M +m)ℓ = mR (this is just Equation 10-4, with origin at the center of the wheel so x 1 = 0 for M, x 2 = R for m, and x cm = ℓ). For small oscillations, the period is given by Equation 15-16 (where M +m is the total mass), therefore T = 2π/ω = 2π I/(M +m)gℓ = 2π× (M +m)R 2 /mgR. Solving for m, we find m = M[(g/R)(T/2π) 2 −1] −1 = (600 g)[(9.8/0.3)× (12/2π) 2 −1] −1 = 5.04 g. Problem 34. A mass m is mounted between two springs of constants k 1 and k 2 , as shown in Fig. 15-41. Show that the angular frequency of oscillation is given by ω = (k 1 +k 2 )/m. figure 15-41 Problem 34. Solution At equilibrium, both springs are either extended or compressed, since their forces must be in opposite directions. If the mass is moved by an amount ∆x to the right of the equilibrium position, the force of the first spring increases by k 1 ∆x to the left, and the force of the second spring decreases by k 2 ∆x to the right (which is also an increase to the left). Thus, the net force is (k 1 +k 2 )∆x to the left, which represents a restoring force (opposite to the displacement ∆x to the right). The effective spring constant is k 1 +k 2 , and angular frequency ω = (k 1 +k 2 )/m for oscillations about the equilibrium position. Problem 35. Repeat the previous problem for the case when the springs are connected as in Fig. 15-42. m k 1 k 2 figure 15-42 Problem 35. Solution The effective spring constant for this combination (see Chapter 5, Problem 73(b)) is k = k 1 k 2 /(k 1 + k 2 ), and ω 2 = k/m. Section 15-4: Circular Motion and Simple Harmonic Motion Problem 36. An object undergoes simple harmonic motion in two mutually perpendicular directions, its position given by r = Asin ωtˆı +Acos ωtˆ. (a) Show that the object remains a fixed distance from the origin—i.e., that its path is circular—and find that distance. (b) Find an expression for the object’s velocity. (c) Show that the speed remains constant, and find its value. (d) What is the angular speed of the object in its circular path? Solution (a) The object’s distance from the origin is |r| = (Asin ωt) 2 + (Acos ωt) 2 = A, a constant, so its path is a circle with that radius. (b) Differentiating, we find v =dr/dt = ωAcos ωtˆı −ωAsin ωtˆ. (Note that v · r = 0, as required for circular motion.) (c) The object’s speed is |v| = (ωAcos ωt) 2 + (−ωAsinωt) 2 = ωA, also a constant. (d) From Equation 12-4, the angular speed is v/r = ωA/A = ω. Problem 37. The equation for an ellipse is x 2 a 2 + y 2 b 2 = 1. Show that two-dimensional simple harmonic motion whose two components have different amplitudes and are π/2 out of phase gives rise to elliptical motion. How are a and b related to the amplitudes? Solution Simple harmonic motions in the x and y directions, with different amplitudes and π/2 out of phase, are x = a cos(ωt +φ) and y = b cos(ωt +φ ±π/2) = ∓b sin(ωt +φ). These describe an elliptical path CHAPTER 15 239 with semi-major or minor axis equal to the amplitudes, a or b, since (x/a) 2 + (y/b) 2 = cos 2 (ωt +φ) + sin 2 (ωt +φ) = 1. Problem 38. The x and y components of motion of a body are harmonic with frequency ratio 1.75 : 1. How many oscillations must each component undergo before the body returns to its initial position? Solution Since ω x /ω y = 1.75/1 = 7/4, seven oscillations in the x direction are completed for four in the y direction. Section 15-5: Energy in Simple Harmonic Motion Problem 39. A 1400-kg car with poor shock absorbers is bouncing down the highway at 20 m/s, executing vertical harmonic motion at 0.67 Hz. If the amplitude of the oscillations is 18 cm, what is the total energy in the oscillations? What fraction of the car’s kinetic energy is this? Neglect rotational energy of the wheels and the fact that not all the car’s mass participates in the oscillation. Solution The total vibrational energy is E vib = 1 2 kA 2 = 1 2 mω 2 A 2 = 1 2 (1400 kg)(2π×0.67 Hz) 2 (0.18 m) 2 = 402 J. The car’s total energy is E = E trans +E vib = 1 2 Mv 2 cm +E vib ≈ 1 2 (1400 kg)(20 m/s) 2 = 280 kJ, so E vib /E is only about 0.144%. Problem 40. A 450-g mass on a spring is oscillating at 1.2 Hz. The total energy of the oscillation is 0.51 J. What is the amplitude of oscillation? Solution The total energy of a one-dimensional mass-spring system is E = 1 2 kA 2 = 1 2 mω 2 A 2 (see Section 15-5), so A = 2E/mω 2 = 2(0.51 J)/(0.45 kg)÷ (2π×1.2 Hz) = 20.0 cm. Problem 41. The motion of a particle is described by x = (45 cm)[sin(πt +π/6)], with x in cm and t in seconds. At what time is the potential energy twice the kinetic energy? What is the position of the particle at this time? Solution The condition that the potential energy equal twice the kinetic energy implies that U(t) = 1 2 kx(t) 2 = 2K(t) = mv(t) 2 , or ωx(t)/v(t) = ± √ 2, where ω = k/m. For x(t) = (45 cm) sin(πt +π/6) as given (note that ω = π(s −1 )), v(t) = dx/dt = ω(45 cm)× cos(πt +π/6), so the above condition becomes tan(πt +π/6) = ± √ 2. There are four angles in each cycle which satisfy this (since tan θ = −tan(π −θ) = tan(π +θ) = −tan(2π −θ)), which are π(t + 1 6 ) = 0.955, 2.19, 4.10, and 5.33 radians. (We chose the cycle with phases between 0 and 2π radians; for any other cycle, an integer multiple of 2π can be added to these angles.) The times corresponding to these phases are t = (0.955/π) − 1 6 = 0.137 s, 0.529 s, 1.14 s, and 1.53 s, respectively. (An integer multiple of 2 can be added to get the times for any other cycle.) The positions of the particle corresponding to these phases are x(0.137 s) = (45 cm) sin(0.955) = 36.7 cm = x(0.529 s) = −x(1.14 s) = −x(1.53 s), since sin θ = sin(π −θ) = −sin(π +θ) = −sin(2π −θ). (During each cycle, the particle passes each of the points ±36.7 cm twice, traveling with the same speed, but in opposite directions.) Problem 42. A torsional oscillator of rotational inertia 1.6 kg·m 2 and torsional constant 3.4 N·m/rad has a total energy of 4.7 J. What are its maximum angular displacement and maximum angular speed? Solution For a torsional oscillator, E = 1 2 κA 2 = 1 2 Iω 2 A 2 = 1 2 I(dθ/dt) 2 max , therefore A = 2E/κ = 2(4.7 J)/(3.4 N·m) = 1.66 rad = 95.3 ◦ , and (dθ/dt) max =ωA= 2E/I = 2(4.7 J)/(1.6 kg·m 2 ) = 2.42 s −1 = 139 ◦ /s. Problem 43. Show that the potential energy of a simple pendulum is proportional to the square of the angular displacement in the small-amplitude limit. Solution The potential energy of a simple pendulum (see Example 8-6 and Fig. 8-14) is U = mgh = mgℓ(1 −cos θ). For small angles, cos θ ≈ 1 − 1 2 θ 2 , so U ≈ 1 2 mgℓθ 2 . 240 CHAPTER 15 Problem 44. The total energy of a mass-spring system is the sum of its kinetic and potential energy: E = 1 2 mv 2 + 1 2 kx 2 . Assuming E remains constant, differentiate both sides of this expression with respect to time and show that Equation 15-4 results. Hint: Remember that v = dx/dt. Solution Since E is a constant, dE dt = 0 = d dt 1 2 kx 2 + 1 2 mv 2 = kx dx dt +mv dv dt . Dividing by v = dx/dt, we find m(dv/dt) = m(d 2 x/dt 2 ) = −kx. Problem 45. A solid cylinder of mass M and radius R is mounted on an axle through its center. The axle is attached to a horizontal spring of constant k, and the cylinder rolls back and forth without slipping (Fig. 15-43). Write the statement of energy conservation for this system, and differentiate it to obtain an equation analogous to Equation 15-4 (see previous problem). Comparing your result with Equation 15-4, determine the angular frequency of the motion. figure 15-43 Problem 45. Solution With reference to Equation 12-24 (and the condition v = ωR for rolling without slipping) K = 1 2 Mv 2 + 1 2 I cm ω 2 = 1 2 Mv 2 + 1 2 ( 1 2 MR 2 )(v/R) 2 = 3 4 Mv 2 . The potential energy of the spring is U = 1 2 kx 2 , where v = dx/dt, so E = K +U = 3 4 M(dx/dt) 2 + 1 2 kx 2 . Differentiating, we find: dE dt =0 = 3 4 M· 2 dx dt d 2 x dt 2 + 1 2 k · 2x dx dt , or d 2 x dt 2 =− 2k 3M x ≡ −ω 2 x. (The energy method is particularly convenient for analyzing small oscillations, since complicated details of the forces can be avoided.) Problem 46. A mass m is free to slide on a frictionless track whose height y as a function of horizontal position x is given by y = ax 2 , where a is a constant with the units of inverse length. The mass is given an initial displacement from the bottom of the track and then released. Find an expression for the period of the resulting motion. Solution The potential energy, relative to the bottom of the track, is U = mgy = mgax 2 (analogous to 1 2 kx 2 ), so we suspect that the x component of the motion is simple harmonic motion with “spring constant” k = 2mga. Then the period is T = 2π m/k = 2π/ √ 2ga . Indeed, F x = −dU/dx = −2mgax = md 2 x/dt 2 represents simple harmonic motion with ω = √ 2ga . (The y component of the motion, however, is not simple harmonic motion.) Section 15-6: Damped Harmonic Motion Problem 47. A 250-g mass is mounted on a spring of constant k = 3.3 N/m. The damping constant for this system is b = 8.4×10 −3 kg/s. How many oscillations will the system undergo during the time the amplitude decays to 1/e of its original value? Solution Since the damping constant is small, the motion is under-damped and Equation 15-20 applies. The time for the amplitude to decay to 1/e of its original value is t = 2m/b = 59.5 s, while the period is T = 2π m/k = 1.73 s, therefore the number of corresponding oscillations is 59.5/1.73 = 34.4. Problem 48. The vibration of a piano string can be described by an equation analogous to Equation 15-20. If the quantity analogous to b/2m in that equation has the value 2.8 s −1 , how long will it take the vibration amplitude to drop to half its original value? Solution From Equation 15-20, one sees that the amplitude is one half the initial value when |x| /A = 1 2 = exp{−(2.8 s −1 )t} or t = (ln 2)/(2.8 s −1 ) = 0.248 s (see Example 15-7). CHAPTER 15 241 Section 15-7: Driven Oscillations and Resonance Problem 49. A mass-spring system has b/m = ω 0 /5, where b is the damping constant and ω 0 the natural frequency. How does its amplitude when driven at frequencies 10% above and below ω 0 compare with its amplitude at ω 0 ? Solution The amplitude at resonance (ω d = ω 0 ) is A res = F 0 ÷ bω 0 , so that Equation 15-23 can be rewritten as: A A res = A (F 0 /bω 0 ) = (bω 0 /m) (ω 2 d −ω 2 0 ) 2 +b 2 ω 2 d /m 2 = ¸ mω 0 b 2 ω 2 d ω 2 0 −1 2 + ω 2 d ω 2 0 ¸ −1/2 If (mω 0 /b) = 5, and (ω d /ω 0 ) = 1.1 (10% above resonance), then A/A res = 1/ 25(1.21 −1) 2 + 1.21 = 65.8%, while for ω d /ω 0 = 0.9 (10% below resonance), A/A res = 1/ 25(0.81 −1) 2 + 0.81 = 76.4%. Problem 50. A car’s front suspension has a natural frequency of 0.45 Hz. The car’s front shock absorbers are worn out, so they no longer provide critical damping. The car is driving on a bumpy road with bumps 40 m apart. At a certain speed, the driver notices that the car begins to shake violently. What speed? Solution The peak amplitude occurs at resonance, when ω d = ω 0, so the car receives an impulse from the bumps once each period. It therefore travels the distance between bumps in one period, or 40 m = vT 0 . Thus v = 40 m/T 0 = (40 m)f 0 = (40 m)(0.45 Hz) = 18 m/s = 64.8 km/h. (Recall f 0 = 1/T 0 .) Problem 51. Show by direct substitution that Equation 15-22 satisfies Equation 15-21 with A given by Equation 15-23. Solution When Equation 15-22 is substituted into Equation 15-21, one obtains m[−ω 2 d Acos(ω d t +φ)] = −kAcos(ω d t +φ) −b[−ω d Asin(ω d t +φ)] + F 0 [cos(ω d t +φ)cos φ + sin(ω d t +φ)sin φ], where we let ω d t = ω d t +φ −φ in the F 0 -term, and used a trigonometric identity. This equation is true if the coefficients of the sin(ω d t +φ) and cos(ω d t +φ) terms on each side are equal, respectively, that is, −mω 2 d A = −kA+F 0 cos φ, and 0 = bω d A +F 0 sin φ. Let ω 2 0 = k/m, and these equations become F 0 cos φ = −m(ω 2 d −ω 2 0 )A, and F 0 sinφ = −bω d A. Squaring and adding, we get Equation 15-23. Problem 52. A harmonic oscillator is underdamped provided that the damping constant b is less than √ 2mω 0 , where ω 0 is the natural frequency of undamped motion. Show that for an underdamped oscillator, Equation 15-23 has a maximum for a driving frequency less than ω 0 . Solution Equation 15-23 for A is a maximum when the denominator of the right-hand side is a minimum. The condition for this is d dω d ¸ (ω 2 d −ω 0 ) 2 +b 2 ω 2 d m 2 = 0 = 2(ω 2 d −ω 2 0 )(2ω d ) + 2ω d b 2 m 2 . Since ω d = 0, we may divide by it, obtaining ω ∗ d = ω 2 0 −b 2 /2m 2 for the frequency of the maximum. Evidently, ω ∗ d < ω 0 . Although the motion is underdamped for b < 2mω 0 , A has a maximum in the physical region (ω d > 0, b > 0) only for b < √ 2mω 0 , and A has sharp resonance-type behavior for b ≪2mω 0 . Paired Problems Problem 53. A particle undergoes simple harmonic motion with amplitude 25 cm and maximum speed 4.8 m/s. Find (a) the angular frequency, (b) the period, and (c) the maximum acceleration. Solution (a) Since v max = ωA, ω = (4.8 m/s)/(0.25 m) = 19.2 s −1 . (b) T = 2π/ω = 0.327 s. (c) a max = ωv max = 92.2 m/s 2 . Problem 54. A particle undergoes simple harmonic motion with maximum speed 1.4 m/s and maximum acceleration 3.1 m/s 2 . Find (a) the angular frequency, (b) the period, and (c) the amplitude of the motion. 242 CHAPTER 15 Solution (a) ω = a max /v max = (3.1 m/s 2 )/(1.4 m/s) = 2.21 s −1 . (b) T = 2π/ω = 2.84 s. (c) A = v max /ω = 63.2 cm. Problem 55. A massless spring of spring constant k = 74 N/m is hanging from the ceiling. A 490-g mass is hooked onto the unstretched spring and allowed to drop. Find (a) the amplitude and (b) the period of the resulting motion. Solution (a) The distance from the initial position of the mass on the unstretched spring, to the equilibrium position, where the net force is zero, is just the amplitude, since the initial velocity for a dropped mass is zero. Then at the equilibrium position, mg = kA, or A = (0.49 kg)× (9.8 m/s 2 )/(74 N/m) = 6.49 cm. (Alternatively, when dropped at the unstretched position (zero spring force), the initial acceleration has its maximum magnitude which is just g, so a max = g = ω 2 A gives the same result, since ω 2 = k/m.) (b) T = 2π m/k = 0.511 s. Problem 56. A massless spring is hanging from the ceiling. A mass is hooked onto the end of the spring and allowed to drop. If the amplitude of the resulting motion is 20 cm, what is its frequency? Solution If the mass is dropped from the unstretched position of the spring, the maximum acceleration has magnitude g (at the top of the oscillations) as mentioned in the solution to Problem 55. Then a max = ω 2 A = g, implies f = ω/2π = g/A/2π = (9.8 m/s 2 )/(0.2 m)/2π = (7/2π) Hz = 1.11 Hz. Problem 57. A meter stick is suspended from one end and set swinging. What is the period of the resulting oscillations, assuming they have small amplitude? Solution The meter stick is a physical pendulum whose CM is ℓ = 0.5 m below the point of suspension through one end. The rotational inertia of the stick about one end is 1 3 M(1 m) 2 , so Equation 15-16 gives the period as T = 2π I/Mgℓ = 2π 2(1 m)/3g = 1.64 s. Problem 58. A meter stick is suspended from a frictionless rod inserted through a small hole at the 25 cm mark. What is the period of small-amplitude oscillations about the stick’s equilibrium position? Solution If the point of suspension of the meter stick in the previous problem is moved so that ℓ = 0.25 m and I = 1 12 M(1 m) 2 +M(0.25 m) 2 = 7 3 M(0.25 m) 2 , the period becomes T = 2π 7(1 m)/12g = 1.53 s. Problem 59. Two balls each of unknown mass m are mounted on opposite ends of a 1.5-m-long rod of mass 850 g. The system is suspended from a wire attached to the center of the rod and set into torsional oscillations. If the torsional constant of the wire is 0.63 N·m/rad and the period of the oscillations is 5.6 s, what is the unknown mass m? Solution The period of a torsional pendulum is given by Equation 15-14, ω = 2π/T = κ/I. The rotational inertia of the rod and two masses, about an axis perpendicular to the rod and through its center, is I = (T/2π) 2 κ = 1 12 Mℓ 2 + 2m( 1 2 ℓ) 2 so m = 1 2 [(T÷ πℓ) 2 κ − 1 3 M] = 1 2 [(5.6 s/π×1.5 m) 2 (0.63 N·m) − 1 3 (0.85 kg)] = 303 g. Problem 60. Figure 15-44 shows a bird feeder that consists of a 340-g solid circular disk 50 cm in diameter suspended by a wire attached at the center. Two 65-g birds land at opposite ends of a diameter, and the system goes into torsional oscillation at 2.6 Hz. What is the torsional constant of the wire? Solution The rotational inertia of the bird feeder (solid disk) and birds (point masses m) about the wire is I = 1 2 MR 2 + 2mR 2 , so solving Equation 15-14 for the torsional constant, we find: κ = (2πf) 2 I = (2π× 2.6 Hz) 2 [ 1 2 (0.34 kg) + 2(0.065 kg)](0.25 m) 2 = 5.00 N·m/rad. Problem 61. Two mass-spring systems with the same mass are undergoing oscillatory motion with the same amplitudes. System 1 has twice the frequency of system 2. How do (a) their energies and (b) their maximum accelerations compare? CHAPTER 15 243 50 cm 340 g figure 15-44 Problem 60. Solution (a) The energy of a mass-spring system is E = 1 2 mω 2 A 2 (see Section 15.5). If m and A are the same, but ω 1 = 2ω 2 , then E 1 = 4E 2 . (b) The maximum acceleration is a max = ω 2 A, so a 1,max = 4a 2,max as well. Problem 62. Two mass-spring systems have the same mass and the same total energy. The amplitude of system 1 is twice that of system 2. How do (a) their frequencies and (b) their maximum accelerations compare? Solution If the mass and total energy of two mass-spring systems is the same, then ω 1 A 1 = ω 2 A 2 . Therefore, A 1 = 2A 2 implies (a) ω 1 = 1 2 ω 2 and (b) a 1,max = ω 2 1 A 1 = ( 1 2 ω 2 ) 2 (2A 2 ) = 1 2 a 2,max . Supplementary Problems Problem 63. While waiting for your plane to take off, you suspend your keys from a thread and set the resulting pendulum oscillating. It completes exactly 90 cycles in 1 minute. You repeat the experiment as the plane accelerates down the runway, and now find the pendulum completes exactly 91 cycles in 1 minute. Find the plane’s acceleration. Solution The solution of Problem 21 shows that the period of a simple pendulum (point mass suspended in a constant gravitational field) in an accelerating reference frame is T = 2π/ω = 2π ℓ/ |g −a 0 |. If a 0 is perpendicular to g (as for the air plane in this problem) then |g − a 0 | = g 2 +a 2 0 , and the circular frequency is ω 2 = g 2 +a 2 0 /ℓ. Since ω 2 0 = g/ℓ at rest, we can eliminate ℓ and solve for a 0 : (ω/ω 0 ) 2 = 1 + (a 0 /g) 2 , or (a 0 /g) = (ω/ω 0 ) 4 −1. The ratio of the frequencies is ω/ω 0 = 91/90, so numerically, a 0 = (9.8 m/s 2 )× (91/90) 4 −1 = 2.08 m/s 2 . Problem 64. A 500-g mass is suspended from a thread 45 cm long that can sustain a maximum tension of 6.0 N before breaking. What is the maximum allowable amplitude for pendulum motion of this system? Solution It is shown in Example 15-4 that the greatest tension in a simple pendulum occurs at the bottom of its swing, where T max = mg(1 +A 2 ), and A is the angular amplitude. For the thread in this problem, T max < 6.0 N, so A < (6.0 N/0.5×9.8 N) −1 = 0.474 (rad) = 27.1 ◦ . (Note that sin 27.1 ◦ = 0.456, about 3.8% less than the angle itself, so this answer is only approximate.) Problem 65. A 500-g block on a frictionless surface is connected to a rather limp spring of constant k = 8.7 N/m. A second block rests on the first, and the whole system executes simple harmonic motion with a period of 1.8 s. When the amplitude of the motion is increased to 35 cm, the upper block just begins to slip. What is the coefficient of static friction between the blocks? Solution If the surfaces of contact are horizontal, it is the frictional force which accelerates the upper block, hence f s = m 2 a(t) ≤ µ s N = µ s m 2 g, or a(t) ≤ µ s g. In simple harmonic motion, a max = ω 2 A, so when the upper block begins to slip, ω 2 A = µ s g, or µ s = (2π/1.8 s) 2 (0.35 m/9.8 m/s 2 ) = 0.44. [Note: the data given in the problem which were not used to find µ s (i.e., k and m 1 ) can be used to calculate that m 2 = 214 g, since ω = 2π/T = k/(m 1 +m 2 ).] Problem 66. The potential energy of a 75-g particle is given by U = ax 2 −bx 4 , where a = 3.5 J/m 2 , b = 1.2 J/m 4 , 244 CHAPTER 15 Problem 65 Solution. and x is in meters. (a) Show that there is a metastable equilibrium at x = 0. (b) Find the frequency of small amplitude oscillations about this equilibrium. Hint: For small x, x 4 ≪x 2 . (c) What is the maximum amplitude for oscillatory (but not necessarily simple harmonic) motion to occur? Solution (a) The equilibrium points (for one-dimensional motion) are solutions of the equation dU/dx = 0 = 2x(a −2bx 2 ). These are x = 0 and x = ± a/2b = ±1.21 m. d 2 U/dx 2 = 2(a −6bx 2 ) is positive for x = 0, which represents a relative minimum of the potential energy and therefore a point of metastable equilibrium. (x = 0 is not an absolute minimum because U(x) < 0 for |x| > a/b, in fact U →−∞ for |x| →∞. d 2 U/dx 2 < 0 for x = ± a/2b, so these points are relative maxima of potential energy, representing unstable equilibria.) (b) In one- dimensional motion, F x = −dU/dx = −2x(a − 2bx 2 ) = md 2 x/dt 2 . For small x, 2bx 2 ≪a so d 2 x/dt 2 ≈ −(2a/m)x, which is the equation for simple harmonic motion with frequency f = 2a/m/2π = 1.54 Hz. (c) The motion is oscillatory as long as the particle is in the potential well surrounding x = 0 (see sketch). This is the case provided the total energy E < U(± a/2b) = a 2 /4b = 2.55 J, and the maximum displacement from the origin, |x| < a/2b = 1.21 m. (See Problems 8-43 and 83, which deal with the negative of this potential energy function, however.) Problem 66 Solution. Problem 67. Repeat Problem 46 for a small solid ball of mass M and radius R that rolls without slipping on the parabolic track. Solution The potential energy of the ball is U = mgy = mgax 2 , and its kinetic energy (rolling without slipping) is (from Equation 12-24) K = 1 2 Mv 2 + 1 2 I cm ω 2 = 1 2 Mv 2 + 1 2 ( 2 5 MR 2 )(v/R) 2 = 7 10 Mv 2 . Since the total mechanical energy, E = U +K, is constant, dE dt = 0 = d dt Mgax 2 + 7 10 Mv 2 = 2Mgax dx dt + 7 5 Mv dv dt . Note that in this problem (and in Problem 46) v = (dx/dt) 2 + (dy/dt) 2 , so the motion is not harmonic, in general. However, for small displacements, v ≈ dx/dt, and d 2 x/dt 2 = −(10ga/7)x, which is simple harmonic motion with period T = 2π/ω = 2π 7/10ga. Problem 68. A child twirls around on a swing, twisting the swing ropes, as shown in Fig. 15-45. As a result, the child and swing rise slightly, with the rise, h, in cm equal to the square of the number of full turns of the swing. When the child stops twisting up the swing, it goes into torsional oscillation. What is the period of this oscillation, assuming that all the potential energy of the system is gravitational? The combined mass of the child and swing is 20 kg, and the rotational inertia of the pair about the appropriate vertical axis is 0.12 kg·m 2 . Solution The number of full turns (revolutions) is θ/2π, so h = (1 cm)(θ/2π) 2 , where θ is the angle (in radians) measured around the appropriate vertical axis. The total energy, E = Mgh + 1 2 Iω 2 = Mg(1 cm)(θ/2π) 2 + 1 2 I(dθ/dt) 2 , is constant, so dE dt = 0 = Mg(1 cm) θ 2π 2 dθ dt +I dθ dt d 2 θ dt 2 , or d 2 θ dt 2 = − Mg(1 cm) 2π 2 I θ. This is simple harmonic motion with period T = 2π 2π 2 I Mg(1 cm) CHAPTER 15 245 = 2π 2 2(0.12 kg·m 2 ) (20 kg)(9.8 m/s 2 )(0.01 m) = 6.91 s. figure 15-45 Problem 68 Solution. Problem 69. A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 23 N/m (Fig. 15-46). The block is oscillating with amplitude 10 cm and with phase constant φ = −π/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together. Determine the frequency, amplitude, and phase constant (relative to the original t = 0) of the resulting motion. 1.2 kg 0.80 kg 1.7 m/s figure 15-46 Problem 69. Solution The simple harmonic motion with just the first block on the spring can be described by Equation 15-9 and the given amplitude and phase constant; x(t) = (10 cm) cos(ω 1 t −π/2) = (10 cm) sin ω 1 t, where ω 1 = k/m 1 = (23 N/m)/(1.2 kg) = 4.38 s −1 . This equation holds up to the time of the collision, i.e., for t < t c , where t c = π/2ω 1 , since for the rightmost point of oscillation, sinω 1 t c = 1, or ω 1 t c = π/2. (This specifies the original zero of time appropriate to the given phase constant of −π/2.) Equation 15-9 also describes the simple harmonic motion after the collision; x(t) = Acos(ω 2 t +φ) for t > t c , where ω 2 = k/(m 1 + m 2 ) = 3.39 s −1 is the angular frequency when both blocks oscillate on the spring (and f 2 = ω 2 /2π = 0.540 Hz, as asked in the problem.) It follows from this that v(t) = −ω 2 A× sin(ω 2 t +φ). The amplitude A and phase constant φ can be determined from these two equations evaluated just after the collision, essentially at t c , if we assume that the collision takes place almost instantaneously; then conservation of momentum during the collision can be applied (see Equation 11-4). Just after the collision, x(t c ) = 10 cm (given) and v(t c ) = (m 1 v 1 + m 2 v 2 )/(m 1 +m 2 ), where just before the collision, v 1 = 0 (given m 1 at rightmost point of its original motion) and v 2 = −1.7 m/s (also given). Numerically, v(t c ) = (−1.7 m/s)(0.8)/(0.8 +1.2) = −68 cm/s. Thus, the two equations become x(t c ) = 10 cm = A× cos(ω 2 t c +φ), and v(t c ) = −68 cm/s = −ω 2 A× sin(ω 2 t c +φ), where ω 2 and t c are known. (In fact, ω 2 t c = ω 2 π/2ω 1 = (π/2) m 1 /(m 1 +m 2 ) = √ 0.6 (π/2) radians = 69.7 ◦ .) Solving for A (using sin 2 +cos 2 = 1), we find A = x(t c ) 2 + [−v(t c )/ω 2 ] 2 = (10 cm) 2 + (68 cm/3.39) 2 = 22.4 cm. Solving for φ (using sin / cos = tan), we find φ = tan −1 [−v(t c )÷ ω 2 x(t c )] −ω 2 t c = tan −1 (68/3.39×10) −69.7 ◦ = −6.22 ◦ = −0.109 radians. (Note: The solution for A is equivalent to calculating the various energies in the second simple harmonic motion, since just after the collision, K(t c ) = 1 2 (m 1 +m 2 )v (t c ) 2 = 1 2 (2 kg)(−0.68 m/s) 2 = 0.462 J, U(t c ) = 1 2 kx(t c ) 2 = 1 2 (23 N/m)(0.1 m) 2 = 0.115 J, E = K(t c ) +U(t c ) = 0.577 J = 1 2 kA 2 , or A = 2(0.577 J)/(23 N/m). Once A is known, φ can also be found from either expression for x(t c ) or v(t c ), e.g., ω 2 t c +φ = cos −1 (10/22.4) = sin −1 (68/3.39×22.4).) Problem 70. The magnitude of the gravitational acceleration inside Earth is given approximately by g(r) = g 0 (r/R E ), where g 0 is the surface value, r is the distance from Earth’s center, and R E is Earth’s radius, the acceleration is directed toward Earth’s center. Suppose a narrow hole were drilled straight through the center of Earth and out the other side. Neglecting air resistance, show that an object dropped into this hole executes simple harmonic motion, and find an expression for the period. Evaluate and compare with the period of a satellite in a circular orbit not far above Earth’s surface. Solution The force of gravity is directed toward the center of the Earth, so an object in this hole (not subject to any 246 CHAPTER 15 other forces) has an acceleration d 2 r/dt 2 = −(g 0 /R E )r. This represents one-dimensional simple harmonic motion about r = 0 (compare to Equation 15-4), with period T = 2π R E /g 0 = 2π (6.37×10 6 m)/(9.81 m/s 2 ) = 84.4 min . Since g 0 = GM E /R 2 E , T = 2π R 3 E /GM E , the same as for a satellite in an Earth-grazing orbit. Problem 71. A small object of mass m slides without friction in a circular bowl of radius R. Derive an expression for small-amplitude oscillations about equilibrium, and compare with that of a simple pendulum. Problem 71 Solution. Solution What is intended in this problem is consideration of just motion in a vertical plane through the axis of the bowl. (This corresponds to the same initial conditions as for a simple pendulum. Other conditions lead to motion like various conical pendulums.) The tangential component of Newton’s second law is −mg sin θ = m(d 2 s/dt 2 ), where s = Rθ. For small angles, −g sin θ ≈ −gθ ≈ R(d 2 θ/dt 2 ). This is simple harmonic motion, with the same equations as for a simple pendulum (see Problem 24), so ω = g/R and T = 2π R/g. Problem 72. A more exact expression than Equation 15-18 for the period of a simple pendulum is T = T 0 ¸ 1 + 1 4 sin 2 1 2 θ 0 + 9 64 sin 4 1 2 θ 0 +· · · , where T 0 = 2π ℓ/g is the period in the limit of arbitrarily small amplitude, and θ 0 is the amplitude. The · · · indicates that additional terms (in fact, infinitely many more) are needed for an exact expression. For a pendulum with T 0 = 1.00 s, plot the period given above versus amplitude for amplitudes from 0 to 45 ◦ . By what percentage does the plotted period differ from T 0 for 30 ◦ and 45 ◦ ? Solution Numerical values are displayed in the table and sketch below. θ 0 1 4 sin 2 ( 1 2 θ 0 ) 9 64 sin 4 ( 1 2 θ 0 ) (T −T 0 )/T 0 15 ◦ .0042593 .0000408 0.4% 30 ◦ .016747 .000631 1.7% 45 ◦ .03661 .00302 4.0% Problem 72 Solution. Problem 73. A mass m is connected between two springs of length L, as shown in Fig. 15-47. At equilibrium, the tension force in each spring is F 0 . Find the period of oscillations perpendicular to the springs, assuming sufficiently small amplitude that the magnitude of the spring tension is essentially unchanged. L L m figure 15-47 Problem 73. Solution Suppose that no forces with components in the direction of motion act on the mass other than the spring forces. If m is given a small displacement CHAPTER 15 247 perpendicular to the springs (as sketched), the net force is F y = −2F 0 sin θ = −2F 0 y/ L 2 +y 2 ≈ −2F 0 y/L, for y ≪L. Newton’s second law gives md 2 y/dt 2 = F y , or d 2 y/dt 2 ≈ −(2F 0 /mL)y. This is the equation for simple harmonic motion with angular frequency ω = 2F 0 /mL and period T = 2π/ω = 2π mL/2F 0 . Problem 73 Solution. Problem 74. A disk of radius R is suspended from a pivot somewhere between its center and edge (Fig. 15-48). For what pivot point will the period of this physical pendulum be a minimum? ? Suspension point figure 15-48 Problem 74 Solution. Solution The period of the physical pendulum shown is T = 2π I Mgℓ = 2π 1 2 MR 2 +Mℓ 2 Mgℓ = 2π ℓ g + R 2 2gℓ . This is a minimum when d dℓ ℓ g + R 2 2gℓ = 0 = 1 g − R 2 2gℓ 2 , or ℓ = R/ √ 2. (The fact that this represents a minimum can be seen either by calculating d 2 T÷ dℓ 2 > 0, or by noting that T →∞ for ℓ →0 and ℓ →∞, and so has a minimum between.) Problem 75. A uniform piece of wire is bent into a V-shape with angle θ between two legs of length ℓ. The wire is placed over a pivot, as shown in Fig. 15-49. Show that the angular frequency of small- amplitude oscillations about this equilibrium is given by ω = 3g cos(θ/2) 2ℓ . Solution The CM of the bent wire is a distance h = (ℓ/2)× cos(θ/2) from the pivot. The rotational inertia of the bent wire (two thin rods) about the pivot is I = 2[ 1 3 ( 1 2 M)ℓ 2 ] = 1 3 Mℓ 2 , where M is the mass of the whole wire. Equation 15-16 for the physical pendulum gives ω = Mgh I = Mg(ℓ/2) cos(θ/2) Mℓ 2 /3 = 3g cos(θ/2) 2ℓ 248 CHAPTER 15 figure 15-49 Problem 75 Solution. Problem 76. Integrating the nonconstant acceleration of a harmonic oscillator over time from the time of maximum displacement to the time of zero displacement should give the velocity at zero displacement. Carry out this integration, using Equation 15-11 for the acceleration, and show that your answer is just the maximum velocity. Solution The time of the maximum (positive) displacement is given by the equation ωt 1 +φ = 0, and the time of the next zero displacement by ωt 2 +φ = π/2, where the displacement is given by Equation 15-9. Thus, t2 t1 a dt = t2 t1 (−ω 2 A) cos(ωt +φ)dt = − ω 2 A ω sin(ωt +φ) t2 t1 = −ωAsin(π/2) +ωAsin(0) = −ωA. This is the maximum velocity, which occurs at zero displacement; it is negative because we assumed the oscillator started at +A. 15-34. A simple model for an automobile suspension consists of a mass attached to a spring. (b) T = 1/f = 1.29 s−1 . (b) and (c) both are represented by equations of the form x = A cos ωt.25 m) = 1. t0 = −φ2 /(2π/T ) = 1 4 T. which is intuitively more obvious. One way this system can oscillate is if the carbon atom stays fixed and the two oxygens move symmetrically on either side of it.6 N/m)(0. frequency 0.694 s. Note that cos(ωt + 2 π) = −sin ωt. 1 1 and phase constant 2 π. If the frequency of this oscillation is 4. period T = 2π/(2.0 s.32 m/s (Equation 15-10).3 kg. Problem 11. Solution We suppose that the other end of the spring is fastened to the spacecraft. (In terms of the period.66×10−27 kg) = 1. φ1 = 0 and ω2 t0 + φ2 = 0.5 s/2π)2 = 63. what is the effective spring constant? The mass of an oxygen atom is 16 u.6 N/m)/(0. the frequency of either oxygen atom is ω = 2πf = k/m. the first at t = 0 and the second at t = t0 .40 Hz. Therefore k = (2π×4×1013 Hz)2 (16×1. what is the astronaut’s mass? Solution (a) This equation has amplitude 15 cm. as suggested in Fig. (c) is nearly the same as (a).19 s (Equation 15-8c).) figure 15-34 Problem 9. Then. If the mass is 1900 kg and the spring constant is 26 kN/m.5 s. . whose mass is much greater than the astronaut’s. Solution With the carbon atom end of either “spring” fixed.70 s.25 m) = 1.232 CHAPTER 15 Solution For a mass on a spring.842 Hz (Equation 15-8b).5t + π/2)]. Equation 15-8a gives ω = k/m = (5. Thus.40 N (Equations 15-2 or 3 or 11).29 s−1 )(0. A simple model of a carbon dioxide (CO2 ) molecule consists of three mass points (the atoms) connected by two springs (electrical forces).51 s (t in seconds is understood).589 Hz. Problem 7. t0 = 1 −φ2 /ω2 = 2 π m/k = 1 π (0. where ω1 = ω2 = k/m (for identical mass-spring systems). Then (a) f = ω/2π = 0. and set into simple harmonic motion. and that the orbiting system is approximately inertial.68×103 N/m.5 s−1 ) = 2. 1 The difference in phase is 2 π = (ω1 t + φ1 ) − (ω2 t + φ2 ) = (ω1 − ω2 )t + (φ1 − φ2 ) = −φ2 . Problem 9. Then Equation 15-8c gives m = k/ω 2 = k(T /2π)2 = (400 N/m)(2. but the amplitude and period of (c) are half those of (b). Solution Suppose that both masses are released from their maximum positive displacements. (b) motion with amplitude 30 cm. (c) vmax = ωA = (5. Except for the difference in phase.2 N/m. Sketch the following simple harmonic motions on the same graph: (a) x = (15 cm)[cos(2. phase constant 0. Both are displaced from equilibrium and the first released at time t = 0. and T = 1/f = 1. and (d) Fmax = mamax = mω 2 A = kxmax = kA = (5. An astronaut in an orbiting spacecraft is “weighed” by being strapped to a spring of constant k = 400 N/m. (c) motion with amplitude 15 cm.43 kg)/(2.2 N/m) = 2 0. Two identical mass-spring systems consist of 430-g masses on springs of constant k = 2. with what frequency and period will the car undergo simple harmonic motion? Problem 10 Solution. phase constant 0.0×1013 Hz. since the phase (the argument of the cosine in Equation 15-9) of each motion is zero at release. period 5. How much later should the second be released so the two oscillations differ in phase by π/2? Solution Equations 15-8b and 8c give f = (1/2π) k/m = 0. Problem 8. If the oscillation period is 2.2 kg) = 5. Problem 10. dx/dt = ωA cos ωt. in terms of given data. and the angular frequency is ω = 2π(32. If we define a = A cos φ and b = A sin φ. The quartz crystal in a digital quartz watch executes simple harmonic motion at 32. (This is 215 Hz. A 50-g mass is attached to a spring and undergoes simple harmonic motion. for the three values given.24 km/s .918 N/m. Therefore. chosen so 15 divisions by 2 give a signal at 1. which is related to the maximum compression of the spring (the amplitude) by v0 = ωA. (b) v0 is the maximum speed. A mass m slides along a frictionless horizontal surface at speed v0 .21 m. If we expand Equation 15-9. Problem 13.0 s.29 s−1 .06 cm/s.) If each face of the crystal undergoes a maximum displacement of 100 nm. After a completely elastic encounter with the spring. An important property of linear differential equations is that a linear combination of solutions is also a solution. Determine (a) the angular frequency. determine (a) how long the mass is in contact with the spring and (b) the maximum compression of the spring. and v0 . we find A cos(ωt + φ) = A(cos φ cos ωt − sin φ sin ωt).817 m (the latter expression.CHAPTER 15 233 Problem 12. 2 (c) A = vmax /ω = amax /ω 2 = vmax /amax = (etc. and (c) the amplitude of the motion.5 m/s) /(15 m/s ) = 0.768 Hz. Solution If x = A sin ωt.248 m/s2 )T 2 = 9. Solution The amplitude is A = 10−7 m. therefore ℓ = (T /2π)2 g = (0. using a trigonometric identity from Appendix A. but it is shown in the paragraph before Equation 15-8a and in the solution to the previous problem that cos ωt and sin ωt are solutions. (b) k = mω 2 = (0. and 3. Substituting into Equation 15-4.93 mm. (c) 2. How long should you make a simple pendulum so its period is (a) 200 ms. m. (b) 5. so it undergoes half a cycle of simple harmonic motion before leaving the spring with speed v0 to the left. respectively. Solution The period and length of a simple pendulum (at the surface of the Earth) are related by Equation 15-18. we find ω = amax /vmax = (15 m/s2 )/(3.768 Hz) = 2. Solution (a) From amax = Aω 2 and vmax = Aω. √ then A = a2 + b2 and φ = tan−1 (b/a). is preferred).) = 2 2 (3. Problem 16. It is easy enough to substitute the given expression into Equation 15-4. find the maximum velocity and acceleration of the crystal faces. vmax = Aω = 2. Solution m v0 figure 15-35 Problem 13. which is satisfied if ω 2 = k/m. Thus A = v0 /ω = v0 m/k.00000 Hz. Show by substitution that x(t) = a cos ωt − b sin ωt is a solution to Equation 15-4. 6.29 s−1 )2 = 0.5 m/s.05 kg)(4. Section 15-3: Applications of Simple Harmonic Motion Problem 17.5 m/s) = 4. and the equivalence of the two forms is evident. and that this form √ is equivalent to Equation 15-9 with A = a2 + b2 and φ = tan−1 (b/a). the mass heads back in the direction it came from. . and d2 x/dt2 = −ω 2 A sin ωt = −ω 2 x.06×105 s−1 . 15-35. Show by substitution that x(t) = A sin ωt is a solution to Equation 15-4.0 min? Problem 14. This takes time equal to half a period 1 T = π/ω = 2 π m/k. Solution (a) While the mass is in contact with the spring. we find m(−ω 2 x) = −kx. Its maximum 2 acceleration is 15 m/s and its maximum speed is 3. and amax = 2 Aω 2 = ωvmax = 4. In terms of k. (b) the spring constant. It strikes a spring of constant k attached to a rigid wall. Problem 15. as shown in Fig. the net horizontal force on it is just the spring force.57 km respectively. Problem 22. and allows it to swing back and forth with small amplitude.) Solution At the “new” equilibrium position. thus T = 2π/ω = 2π A/g = 2π 0. A mass is attached to a vertical spring. At the heart of a grandfather clock is a simple pendulum 1. the low point is 5. This is also the maximum acceleration during the simple harmonic motion (since amax occurs where the displacement is maximum).8 m/s ) = 4.029 m/(9.24×103 s = 70. mg = kx1 (see the beginning of Section 15-3) so m/k = x1 /g = 0. (b) accelerating upward with acceleration a = 1 g. T = ∞ (there is no restoring torque and the pendulum does not oscillate). (d) in free fall? Solution At the highest point.8 cm below this.3 m÷ (9.21 s. A 640-g hollow ball 21 cm in diameter is suspended by a wire and is undergoing torsional oscillations at a frequency of 0. (d) If a0 = g. Then Newton’s F = m(a + a0 ). while the right-hand side. or F − ma0 = ma. What is the period of oscillation? Problem 21.8 cm. can be combined with the torque of gravity. What is its period if the rocket is (a) at rest on its launch pad. 3 Equation 15-14 gives κ = ω 2 I = (2π×0. which is 30 cm below the spring’s original equilibrium. if we replace g by |g − a0 |.21 m)2 = 0.78 Hz)2 × 2 ( 3 ×0. (a) If a0 = 0. r × mg.113 N·m/rad. For small oscillations about the equilibrium position (which depends on a0 ). is the same as Equation 15-15. A 340-g mass is attached to a vertical spring and lowered slowly until it rests at a new equilibrium position.) Let a be the acceleration of the pendulum relative to the rocket.342 s. (b) Take the y axis positive up. 1 Then a0 = 2 gˆ and g = −gˆ so T = 2π  . Thus T = 1. Solution (It may be helpful to think of an elevator instead of a rocket in this problem. The length of the lecture (if we assume that the student became bored the instant it began) is 6279(2π)× 2(0.17 m)/3(9.10 s. The peak-to-peak displacement is 2A = 5. bored by a lecture on simple harmonic motion. length 17 cm) by the tip with his frictionless fingers.64 kg)( 1 ×0. so the 1 time between ticks is 2 T = π ℓ/g = 1. 2 Problem 23. the spring is in the original unstretched equilibrium position it had before the mass was attached. A physics student.7 min . so the acceleration is just g (downward). 2ℓ/3g.8 m/s2 ) = 0. Problem 18.45 m long. we find the period is T = 2π 2ℓ/3 g. as before. What is the period of the motion? Problem 20. The system is then set into simple harmonic motion. how long does the lecture last? (Hint: See Example 15-5. the clock ticks each time the pendulum reaches its maximum displacement in either direction.2 g. What is the time interval between ticks? Solution The clock ticks twice each period of oscillation. so amax = g = ω 2 A. (c) accelerating downward 2 1 with acceleration a = 2 g. A pendulum of length ℓ is mounted in a rocket. At the high point of the oscillation. and let a0 be the acceleration of the rocket relative to the ground . second law is The rotational analog of this equation is the appropriate generalization of Equation 15-15 for a simple pendulum in an accelerating frame. Problem 19. What is the torsional constant of the wire? ℓ/(g + 1 g) = 2 ℓ/(g − 1 g) = 2 2π Solution The rotational inertia of a hollow sphere about an axis through its center is 2 MR 2 (see Table 12-2). |r × ma| = Iα. Solution Regarding the pencil as a thin rod suspended from one end. T = 2π ℓ/g.8 m/s2 ) = (T /2π)2 . (c) If a0 = − 1 gˆ T = 2π 2 . as in Example 15-5.234 CHAPTER 15 (assumed to be an inertial system). the period is T = 2π ℓ/ |g − a0 |.78 Hz. If it completes 6279 full cycles during the lecture. The “fictitious” torque (about the point of suspension). r × (−ma0 ). there is no spring force (since the spring is unstretched). 2 2π 2ℓ/g. which then goes into oscillation. idly picks up his pencil (mass 9. rather than the angular displacement and torque.CHAPTER 15 235 Problem 24. through Equation 15-14. Two 75-kg steelworkers leap onto opposite ends of the beam. Problem 24 Solution. or M = 6(75 kg)(0. In terms of k and κ. For small displacements. Solution The frequencies before and after the steel workers jump. A solid disk of radius R is suspended from a spring of linear spring constant k and torsional constant κ. and after. can be related to the rotational inertias of the beam. I2 = I1 + 2m(ℓ/2)2 = 12 ℓ2 × (M + 6m). f2 /f1 = I1 /I2 . ≈ m 2 = −mg . m figure 15-36 Problem 25.8) ℓ (M + 6m). as shown in Fig. Problem 27. Problem 25. x ≈ ℓθ = s. 15-36.8f1 . for a simple pendulum of mass m and length ℓ. (The radial component guarantees that the motion follows a circular arc. 15-37. Before. or = − dt2 dt ℓ dt2 ℓ This is the equation for simple harmonic motion with period T = 2π/ω = 2π ℓ/g. If the frequency . f2 = 0. we 1 find k/M = κ/I = κ/( 2 M R2 ). of torsional oscillations diminishes by 20%. If g is to be measured to within 1 mgal(1 gal = 1 cm/s2 ) and if the period can be measured with arbitrary accuracy. I1 = 12 M ℓ2 1 (Table 12-2).) The horizontal displacement is x = ℓ sin θ. with and without them. Derive the period of a simple pendulum by considering the horizontal displacement x and the force acting on the bob. since the torsional 1 constant κ is not changed. what value of R will give the same period for the vertical and torsional oscillations of this system? Solution Equating the angular frequencies for vertical and torsional oscillations (Equations 15-8a and 15-14). or R = 2κ/k. how accurately must the length of the pendulum be known? Problem 26. so the equation of motion is approximately d2 x x d2 x g d2 s x. Solution The tangential component of Newton’s second law. what is the mass of the beam? figure 15-37 Problem 26. Therefore I1 = (f2 /f1 )2 I2 = 12 M ℓ2 = 1 2 2 12 (0.0 m long is suspended from a crane and is undergoing torsional oscillations. where m = 75 kg is each steelworker’s 1 mass. is m(d2 s/dt2 ) = −mg sin θ. A thin steel beam 8.64) = 800 kg. A particular gravimeter uses the period of a 1-m-long pendulum to determine g. as shown in Fig.64)÷ (1 − 0. Geologists use an instrument called a gravimeter to measure the local acceleration of gravity. 0 cm in diameter. This is the equation for simple harmonic motion with ω = ( 1 kR2 )/I = 4 ( 1 kR2 )/( 1 M R2 ) = 4 2 k/2M. The restoring torque of 2 1 the spring about the central axle. as shown in Fig. Problem 29. s. . Problem 28 Solution. ℓ = (T /2π) g. + mℓ2 /mgℓ Problem 30. equals I(d θ/dt ) for the disk. 15-39. A spring of spring constant k is connected to the disk at a point 1 R above the axle..0 cm long. .875 m/9. suspended by an essentially massless string 80. and in equilibrium 2 runs horizontally to a wall.147%. = Tsimp = Tsimp (2R2 /5ℓ2 ) 1 + 0. 15-38. A thin. 2 Problem 28. Solution For a simple pendulum. Hint: You may find the parallel-axis theorem useful.e. figure 15-38 Problem 29 Solution.236 CHAPTER 15 oscillations is 2π 2R/g. then the fractional errors in ℓ and g are the same. Solution For small (approximately horizontal) displacements of the end of the spring. sin θ ≃ θ and cos θ ≃ 1.)Tsimp = 1.5)2 = (1. where we used the parallel axis theorem for I. The fractional error is (Tphys − Tsimp )/Tphys = 0. x ≈ 1 Rθ. with ℓ = h = R. The more exact expression for this type of physical pendulum (called Borda’s pendulum) is Tphys = 2π I/mgℓ = 2π 1+ 2 2 5 mR Solution Equation 15-16 gives T = 2π I/M gℓ = 2π (M R2 + M R2 )/M gR = 2π 2R/g. How much error is introduced by the simple pendulum approximation? Hint: Remember the parallel-axis theorem. what is the angular frequency of the resulting oscillations? Hint: For small θ.00147 . If the period is known precisely (∆T ≈ 0). . Tsimp = 2π ℓ/g = 2π 0.4(7. Calculate the period of this pendulum.8 m/s2 = 1. . s. A pendulum consists of a 320-g solid ball 15. Show that the period of the figure 15-39 Problem 30 Solution. If the disk is rotated slightly away from equilibrium. ∆ℓ/ℓ = ∆g/g = 1 mgal÷ 980 gal ≈ 10−6 . uniform hoop of mass M and radius R is suspended from a thin horizontal rod and set oscillating with small amplitude. treating it first as a simple pendulum and then as a physical pendulum. i. . as shown in Fig.5/87.880 . Solution For a simple pendulum. τ ≈ −( 2 R)(kx) ≈ 1 2 2 2 − 4 kR θ. .877 . A solid disk of mass M and radius R is mounted on a horizontal axle. . The unbalanced torque about the point of contact. sin θ ≈ θ. 2mg/3M R (see also Problem 33. This result is not the same as that of the previous problem. the kinetic energy is approximately that of just the disk. Solution An exact solution of this problem requires advanced methods in analytical mechanics. Alternatively. Repeat the previous problem for the case when the disk does not contact the ground but is mounted on a frictionless horizontal axle through its center. By analogy 1 with a mass-spring system (where E = 2 kx2 + 1 2 2 m(dx/dt) ). Why is your answer different? Solution The net (restoring) torque about the axle is due only to the point mass. This is a restoring torque because it is in the opposite direction to the angular displacement. whose only irregularity is the presence of the tire valve stem. 15-40). which is the equation for simple harmonic motion with period T = 2π I/mgR = 2π (R/g)(1 + M/2m). rolling without slipping.). For small angles. whereas the answer to Problem 31 holds only for m ≪ M. or I = 1 M R2 + M R2 = 2 M R2 (recall 2 Equation 12-20). i. in this problem there is no restriction on the relative size of the masses. (Also. so U = mgy = mgR(1 − cos θ) ≈ 1 mgRθ2 . Show that the period of this motion is given by T = 2π 3M R/2mg. she notices the front wheel executing a slow. because the forces acting are different (there is an axle. there are two simple heuristic arguments showing that the period given is approximately correct. so α ≈ − mgRθ/I = −(2mg/3M R)θ. 2 For m ≪ M. The total energy is E = 1 3 1 2 K + U ≈ 2 (mgR)θ + 2 ( 2 M R2 )(dθ/dt)2 . The disk is rolled slightly away from its equilibrium position and released. as before. It rolls back and forth without slipping. Considering the wheel to be a thin ring of mass 600 g and radius 30 cm. O (which is instantaneously at rest for rolling without slipping) is τ = Iα = −mgR sin θ ≈ −mgRθ. as shown in the sketch. the rotational inertia is just that of the disk 3 about O. If we neglect m compared to M. θ ≪ 1. for small θ. this represents simple harmonic motion with angular frequency (analogous to k/m) of 3 (mgR)/( 2 M R2 ) = Problem 44). However. After she gets it upside down. determine the mass of the valve stem. A point mass m is attached to the rim of an otherwise uniform solid disk of mass M and radius R (Fig. Problem 32. so α = −(mgR/I)θ. we could use Equation 15-16 for a physical pendulum. This is the equation for simple harmonic motion (see Equation 15-13) with period T = 2π/ω = 2π× 3M R/2mg (see Equation 15-14). for small displacements from equilibrium. m ≪ M.e. if the point mass is also assumed to be much smaller than the mass of the disk. whose CM is suspended a distance ℓ = mR/(m + M ) from the axle.) Problem 32 Solution. from equilibrium. therefore τ = −mgR sin θ = Iα. . or K = 1 1 3 2 2 2 2 Iω = 2 ( 2 M R )(dθ/dt) . 870 figure 15-40 Problem 31 Solution. there is no friction.. small-amplitude back-and-forth rotational motion with a period of 12 s. the potential energy varies like that of just the point mass (since that of the disk is constant when it rolls on a horizontal surface). etc. A cyclist turns her bicycle upside down to tinker with it. Consider the disk when it is displaced by a small angle. 1 where I = 2 M R2 + mR2 . Alternatively.CHAPTER 15 237 Problem 31. Then T = 2π I/(m + M )gℓ . The equation for an ellipse is y2 x2 + 2 = 1.) (c) The object’s speed is |v| = (ωA cos ωt)2 + (−ωA sin ωt)2 = ωA. therefore T = 2π/ω = 2π I/(M + m)gℓ = 2π× (M + m)R2 /mgR . and the force of the second spring decreases by k2 ∆x to the right (which is also an increase to the left).8/0. since their forces must be in opposite directions. The effective spring constant is k1 + k2 . with origin at the center of the wheel so x1 = 0 for M.. as required for circular motion. 15-41. (c) Show that the speed remains constant. These describe an elliptical path . the force of the first spring increases by k1 ∆x to the left. If the mass is moved by an amount ∆x to the right of the equilibrium position. Solution At equilibrium. the period is given by Equation 15-16 (where M + m is the total mass). with rotational inertia I = M R2 + mR2 about its central axle. Thus. (d) From Equation 12-4. also a constant. both springs are either extended or compressed. Repeat the previous problem for the case when the springs are connected as in Fig. that its path is circular—and find that distance. 0. A mass m is mounted between two springs of constants k1 and k2 . Problem 37. so its path is a circle with that radius. and find its value. where M = 600 g is the mass of the wheel (thin ring) and m is the mass of the valve stem (a circumferential point mass). its position given by r = A sin ωtˆ + A cos ωtˆ (a) Show that ı . we find m = M [(g/R)(T /2π)2 − 1]−1 = (600 g)[(9. (d) What is the angular speed of the object in its circular path? Solution (a) The object’s distance from the origin is |r| = (A sin ωt)2 + (A cos ωt)2 = A.e. the angular speed is v/r = ωA/A = ω. figure 15-41 Problem 34. Solution The effective spring constant for this combination (see Chapter 5. (b) Differentiating. x2 = R for m. Problem 34. Problem 73(b)) is k = k1 k2 /(k1 + k2 ). and ω 2 = k/m. An object undergoes simple harmonic motion in two mutually perpendicular directions. and xcm = ℓ). we find v = dr/dt = ωA cos ωtˆ − ωA sin ωtˆ (Note that v · r = ı . k1 k2 m figure 15-42 Problem 35. Solution Simple harmonic motions in the x and y directions. with different amplitudes and π/2 out of phase. the net force is (k1 + k2 )∆x to the left. Solving for m.04 g. which represents a restoring force (opposite to the displacement ∆x to the right). How are a and b related to the amplitudes? Problem 35. as shown in Fig.3)× (12/2π)2 − 1]−1 = 5. are x = a cos(ωt + φ) and y = b cos(ωt + φ ± π/2) = ∓b sin(ωt + φ). For small oscillations. Show that the angular frequency of oscillation is given by ω= (k1 + k2 )/m. a constant. the object remains a fixed distance from the origin—i. and angular frequency ω = (k1 + k2 )/m for oscillations about the equilibrium position. 2 a b Show that two-dimensional simple harmonic motion whose two components have different amplitudes and are π/2 out of phase gives rise to elliptical motion.238 CHAPTER 15 Solution The bicycle wheel may be regarded as a physical pendulum. (b) Find an expression for the object’s velocity. Section 15-4: Circular Motion and Simple Harmonic Motion Problem 36. 15-42. The distance of the CM from the axle is given by (M + m)ℓ = mR (this is just Equation 10-4. 10.7 cm = x(0. and 1. Section 15-5: Energy in Simple Harmonic Motion Problem 39. E = 2 κA2 = 1 Iω 2 A2 = 2 1 2 2E/κ = 2 I(dθ/dt)max . with x in cm and t in seconds.66 rad = 95.4 N·m) = 1. The car’s total energy is E = Etrans + Evib = 1 1 2 2 2 M vcm + Evib ≈ 2 (1400 kg)(20 m/s) = 280 kJ. Problem 43. so Evib /E is only about 0. what is the total energy in the oscillations? What fraction of the car’s kinetic energy is this? Neglect rotational energy of the wheels and the fact that not all the car’s mass participates in the oscillation. A 450-g mass on a spring is oscillating at 1. What is the amplitude of oscillation? Solution The total energy of a one-dimensional mass-spring 1 system is E = 1 kA2 = 2 mω 2 A2 (see Section 15-5).955. (An integer multiple of 2 can be added to get the times for any other cycle.529 s. .3◦ .14 s. What are its maximum angular displacement and maximum angular speed? Solution 1 The total vibrational energy is Evib = 2 kA2 = 1 1 2 2 2 2 2 mω A = 2 (1400 kg)(2π×0. cos θ ≈ 1 − 1 θ2 . therefore A = 2(4. so the above condition becomes √ tan(πt + π/6) = ± 2.42 s−1 = 139◦/s. (During each cycle.137 s) = (45 cm) sin(0.14 s) = −x(1. Problem 42. For small angles. traveling with the same speed. the particle passes each of the points ±36. Problem 40.) Problem 38.4 N·m/rad has a total energy of 4.18 m) = 402 J. How many oscillations must each component undergo before the body returns to its initial position? Solution Since ωx /ωy = 1. A torsional oscillator of rotational inertia 1. since sin θ = sin(π − θ) = −sin(π + θ) = −sin(2π − θ). The motion of a particle is described by x = (45 cm)[sin(πt + π/6)].51 J. At what time is the potential energy twice the kinetic energy? What is the position of the particle at this time? Solution The potential energy of a simple pendulum (see Example 8-6 and Fig.53 s. v(t) = dx/dt = ω(45 cm)× cos(πt + π/6). a or b. A 1400-kg car with poor shock absorbers is bouncing down the highway at 20 m/s. Problem 41.529 s) = −x(1.955/π) − 6 = 0. executing vertical harmonic motion at 0.33 radians. and (dθ/dt)max = ωA = 2E/I = 2(4. but in opposite directions. Solution 1 For a torsional oscillator.955) = 36. The x and y components of motion of a body are harmonic with frequency ratio 1. (We chose the cycle with phases between 0 and 2π radians.75/1 = 7/4.7 cm twice.51 J)/(0.) The times corresponding to these 1 phases are t = (0. which are π(t + 6 ) = 0.CHAPTER 15 with semi-major or minor axis equal to the amplitudes.7 J)/(3. 2. 239 Solution The condition that the potential energy equal twice 1 the kinetic energy implies that U (t) = 2 kx(t)2 = √ 2K(t) = mv(t)2 . For x(t) = (45 cm) sin(πt + π/6) as given (note that ω = π(s−1 )). so 2 A = 2E/mω 2 = 2(0. respectively. since (x/a)2 + (y/b)2 = cos2 (ωt + φ) + sin2 (ωt + φ) = 1. 4.67 Hz.) The positions of the particle corresponding to these phases are x(0.7 J)/(1.19. or ωx(t)/v(t) = ± 2.45 kg)÷ (2π×1.144%. Show that the potential energy of a simple pendulum is proportional to the square of the angular displacement in the small-amplitude limit.7 J. 0.6 kg·m2 ) = 2. If the amplitude of the oscillations is 18 cm.6 kg·m2 and torsional constant 3. The total energy of the oscillation is 0. There are four angles in each cycle which satisfy this (since tan θ = − tan(π − θ) = 1 tan(π + θ) = − tan(2π − θ)).2 Hz. for any other cycle.2 Hz) = 20.67 Hz) (0.75 : 1. and 5. seven oscillations in the x direction are completed for four in the y direction.0 cm. where ω = k/m. 1. an integer multiple of 2π can be added to these angles. 2 1 so U ≈ 2 mgℓθ2 . 8-14) is U = mgh = mgℓ(1 − cos θ).53 s).137 s. figure 15-43 Problem 45.248 s (see Example 15-7). one sees that the amplitude is one half the initial value when |x| /A = 1 = 2 exp{−(2. (The y component of the motion.8 s−1 )t} or t = (ln 2)/(2. A solid cylinder of mass M and radius R is mounted on an axle through its center. Then the period is T = 2π m/k = 2π/ 2ga .5/1. The 1 2 potential energy of the spring is U = 2 kx . determine the angular frequency of the motion.3 N/m. since complicated details of the forces can be avoided.73 = 34. 15-43). the motion is under-damped and Equation 15-20 applies. If the quantity analogous to b/2m in that equation has the value 2. The axle is attached to a horizontal spring of constant k. however. .240 CHAPTER 15 Problem 44. differentiate both sides of this expression with respect to time and show that Equation 15-4 results. we find: dE 1 d2 x 3 dx dx . relative to the bottom of the track.) Problem 48. Write the statement of energy conservation for this system. Hint: Remember that v = dx/dt. or + k · 2x =0 = M ·2 dt 4 dt dt2 2 dt d2 x 2k =− x ≡ −ω 2 x. Problem 46. A mass m is free to slide on a frictionless track whose height y as a function of horizontal position x is given by y = ax2 .4. and the cylinder rolls back and forth without slipping (Fig.) Dividing by v = dx/dt. where v = dx/dt.73 s. Assuming E remains constant. How many oscillations will the system undergo during the time the amplitude decays to 1/e of its original value? Solution Since the damping constant is small. dE d 1 2 1 =0= kx + mv 2 dt dt 2 2 = kx dx dv + mv . The damping constant for this system is b = 8. The time for the amplitude to decay to 1/e of its original value is t = 2m/b = 59. Find an expression for the period of the resulting motion. is U = mgy = mgax2 (analogous to 1 kx2 ). Comparing your result with Equation 15-4. how long will it take the vibration amplitude to drop to half its original value? Solution From Equation 15-20. where a is a constant with the units of inverse length.4×10−3 kg/s. Fx = −dU/dx = −2mgax = md2 x/dt2 represents simple harmonic motion with √ ω = 2ga . Section 15-6: Damped Harmonic Motion Problem 47. while the period is T = 2π m/k = 1.8 s−1 ) = 0. and differentiate it to obtain an equation analogous to Equation 15-4 (see previous problem). we find m(dv/dt) = m(d2 x/dt2 ) = −kx. 4 2 Differentiating. Indeed. A 250-g mass is mounted on a spring of constant k = 3. so E = K + U = 3 M (dx/dt)2 + 1 kx2 . Solution Since E is a constant. The vibration of a piano string can be described by an equation analogous to Equation 15-20. dt dt Solution The potential energy.5 s. The mass is given an initial displacement from the bottom of the track and then released. so 2 we suspect that the x component of the motion is simple harmonic motion with “spring constant” k =√ 2mga. dt2 3M (The energy method is particularly convenient for analyzing small oscillations. Solution With reference to Equation 12-24 (and the condition v = ωR for rolling without slipping) K = 1 M v 2 + 2 1 1 1 1 3 2 2 2 2 2 2 Icm ω = 2 M v + 2 ( 2 M R )(v/R) = 4 M v . The total energy of a mass-spring system is the sum of its kinetic and potential energy: E = 1 1 2 2 2 mv + 2 kx .8 s−1 . therefore the number of corresponding oscillations is 59. Problem 45. is not simple harmonic motion. 4%. or 40 m = vT0 .8%. we get Equation 15-23. b > 0) only for b < 2mω0 . we may divide by it. Solution The amplitude at resonance (ωd = ω0 ) is Ares = F0 ÷ bω0 . A harmonic oscillator is underdamped provided √ that the damping constant b is less than 2mω0 . then A/Ares = 1/ 25(1. (Recall f0 = 1/T0 . What speed? Solution The peak amplitude occurs at resonance. and (c) the maximum acceleration. (c) amax = ωvmax = 92. m2 ∗ Since ωd = 0. and A has sharp resonance-type behavior for b ≪ 2mω0 . (b) the period. Show by direct substitution that Equation 15-22 satisfies Equation 15-21 with A given by Equation 15-23. A particle undergoes simple harmonic motion with maximum speed 1. Let 2 ω0 = k/m.25 m) = 19. respectively.2 m/s2 . A has a maximum in the √ physical region (ωd > 0. while for ωd /ω0 = 0. This equation is true if the . Problem 54.21 = 65. A mass-spring system has b/m = ω0 /5. Although the motion is underdamped for b < 2mω0 . and F0 sin φ = −bωd A. The car is driving on a bumpy road with bumps 40 m apart. If (mω0 /b) = 5. The car’s front shock absorbers are worn out. where b is the damping constant and ω0 the natural frequency. where ω0 is the natural frequency of undamped motion. Solution When Equation 15-22 is substituted into 2 Equation 15-21.45 Hz.8 m/s.4 m/s and maximum 2 acceleration 3.1 m/s . A particle undergoes simple harmonic motion with amplitude 25 cm and maximum speed 4. Solution (a) Since vmax = ωA. one obtains m[−ωd A cos(ωd t + φ)] = −kA cos(ωd t + φ) − b[−ωdA sin(ωd t + φ)] + F0 [cos(ωd t + φ)cos φ + sin(ωd t + φ)sin φ]. and used a trigonometric identity. the driver notices that the car begins to shake violently. (b) the period.45 Hz) = 18 m/s = 64. A/Ares = 1/ 25(0. A car’s front suspension has a natural frequency of 0. that is.21 − 1)2 + 1. and (ωd /ω0 ) = 1. so the car receives an impulse from the bumps once each period. Find (a) the angular frequency. so they no longer provide critical damping.CHAPTER 15 241 Section 15-7: Driven Oscillations and Resonance Problem 49. −mωd A = −kA + F0 cos φ. It therefore travels the distance between bumps in one period. (b) T = 2π/ω = 0.9 (10% below resonance). ωd < ω0 . and these equations become F0 cos φ = 2 2 −m(ωd − ω0 )A.81 = 76. ω = (4.8 km/h.2 s−1 . Show that for an underdamped oscillator. The condition for this is 2 2 d (ωd − ω0 )2 + b2 ωd 2 dωd m 2 2 2(ωd − ω0 )(2ωd ) + 2ωd b2 =0= . Problem 51. where we let ωd t = ωd t + φ − φ in the F0 -term. and 0 = bωd A + F0 sin φ. Find (a) the angular frequency.8 m/s)/(0. ω0 ∗ Evidently. Problem 50.1 (10% above resonance). obtaining ωd = 2 − b2 /2m2 for the frequency of the maximum. At a certain speed. Problem 52.327 s. when ωd = ω0.) Paired Problems Problem 53. How does its amplitude when driven at frequencies 10% above and below ω0 compare with its amplitude at ω0 ? coefficients of the sin(ωd t + φ) and cos(ωd t + φ) terms 2 on each side are equal. Equation 15-23 has a maximum for a driving frequency less than ω0 . Thus v = 40 m/T0 = (40 m)f0 = (40 m)(0. so that Equation 15-23 can be rewritten as: A A = = Ares (F0 /bω0 ) = mω0 b 2 (bω0 /m) 2 2 2 (ωd − ω0 )2 + b2 ωd /m2 2 ωd 2 −1 ω0 2 ω2 + d 2 ω0 −1/2 Solution Equation 15-23 for A is a maximum when the denominator of the right-hand side is a minimum. and (c) the amplitude of the motion.81 − 1)2 + 0. Squaring and adding. mg = kA. where the net force is zero.21 s (b) T = 2π/ω = 2.63 N·m) − 1 3 (0. (c) A = vmax /ω = 63. 58. How do (a) their energies and (b) their maximum accelerations compare? .8 m/s )/(74 N/m) = 6. Figure 15-44 shows a bird feeder that consists of a 340-g solid circular disk 50 cm in diameter suspended by a wire attached at the center.8 m/s2 )/(0.34 kg) + 2(0.6 s/π×1. assuming they have small amplitude? Solution The rotational inertia of the bird feeder (solid disk) and birds (point masses m) about the wire is I = 1 2 2 2 M R + 2mR .) (b) T = 2π m/k = 0. about an axis perpendicular to the rod and through its center. is 1 1 I = (T /2π)2 κ = 12 M ℓ2 + 2m( 2 ℓ)2 so m = 1 [(T ÷ 2 1 1 2 πℓ) κ − 3 M ] = 2 [(5. A mass is hooked onto the end of the spring and allowed to drop.2 cm. since the initial velocity for a dropped mass is zero.49 cm. If the torsional constant of the wire is 0. we find: κ = (2πf )2 I = (2π× 2 1 2.1 m/s )/(1. A 490-g mass is hooked onto the unstretched spring and allowed to drop. 2 −1 Problem .25 m)2 = 3 M (0. A massless spring is hanging from the ceiling.64 s. and the system goes into torsional oscillation at 2.85 kg)] = 303 g. ω = 2π/T = κ/I. implies f = ω/2π = g/A/2π = (9. the period becomes T = 2π 7(1 m)/12g = 1. so Equation 15-16 gives the period as 3 T = 2π I/M gℓ = 2π 2(1 m)/3g = 1.5-m-long rod of mass 850 g. System 1 has twice the frequency of system 2.25 m and 7 1 I = 12 M (1 m)2 + M (0. so amax = g = ω 2 A gives the same result. A meter stick is suspended from a frictionless rod inserted through a small hole at the 25 cm mark.5 m)2 (0. What is the torsional constant of the wire? Problem 57.49 kg)× 2 (9. Solution If the point of suspension of the meter stick in the previous problem is moved so that ℓ = 0. Then amax = ω 2 A = g.4 m/s) = 2. what is its frequency? Solution If the mass is dropped from the unstretched position of the spring. Two balls each of unknown mass m are mounted on opposite ends of a 1. Solution (a) The distance from the initial position of the mass on the unstretched spring. The rotational inertia of the stick about one end is 1 M (1 m)2 .25 m)2 = 5. Find (a) the amplitude and (b) the period of the resulting motion. If the amplitude of the resulting motion is 20 cm. the maximum acceleration has magnitude g (at the top of the oscillations) as mentioned in the solution to Problem 55. Problem 59. A massless spring of spring constant k = 74 N/m is hanging from the ceiling. What is the period of the resulting oscillations. to the equilibrium position. The system is suspended from a wire attached to the center of the rod and set into torsional oscillations. (Alternatively.242 CHAPTER 15 Solution (a) ω = amax /vmax = (3. What is the period of small-amplitude oscillations about the stick’s equilibrium position? Problem 55. or A = (0.6 Hz. so solving Equation 15-14 for the torsional constant.00 N·m/rad. Two mass-spring systems with the same mass are undergoing oscillatory motion with the same amplitudes. when dropped at the unstretched position (zero spring force). Two 65-g birds land at opposite ends of a diameter.53 s.2 m)/2π = (7/2π) Hz = 1.84 s. A meter stick is suspended from one end and set swinging. the initial acceleration has its maximum magnitude which is just g.25 m)2 .6 Hz) [ 2 (0. what is the unknown mass m? Solution The period of a torsional pendulum is given by Equation 15-14.63 N·m/rad and the period of the oscillations is 5. Problem 56. Problem 61. The rotational inertia of the rod and two masses.5 m below the point of suspension through one end.6 s.511 s. since ω 2 = k/m. Problem 60. Then at the equilibrium position.11 Hz. Solution The meter stick is a physical pendulum whose CM is ℓ = 0.065 kg)](0. is just the amplitude. 35 m/9. 340 g 50 cm Problem 64. we can eliminate ℓ 0 and solve for a0 : (ω/ω0 )2 = 1 + (a0 /g)2 . While waiting for your plane to take off. amax = ω 2 A.max as well.8 s. and the whole system executes simple harmonic motion with a period of 1. or (a0 /g) = (ω/ω0 )4 − 1. [Note: the data given in the problem which were not used to find µs (i. Tmax < 6. k and m1 ) can be used to calculate that m2 = 214 g. so numerically. since ω = 2π/T = k/(m1 + m2 ). .max ..0 N/0. Therefore. Supplementary Problems Problem 63.456. What is the coefficient of static friction between the blocks? Solution If the mass and total energy of two mass-spring systems is the same. where a = 3. (Note that sin 27. Solution If the surfaces of contact are horizontal. and the circular frequency is ω 2 = 0 2 g 2 + a2 /ℓ.max = 2 1 1 2 ω1 A1 = ( 2 ω2 )2 (2A2 ) = 2 a2. so when the upper block begins to slip. or a(t) ≤ µs g. You repeat the experiment as the plane accelerates down the runway. (b) The maximum acceleration is amax = ω 2 A. For the thread in this problem.5 J/m2 .5×9. about 3. the upper block just begins to slip.8 N) − 1 = 0.7 N/m.2 J/m4 . ω 2 A = µs g.max = 4a2. If m and A are the same.0 N before breaking. How do (a) their frequencies and (b) their maximum accelerations compare? Problem 65.8 m/s ) = 0.8 s)2 (0. hence fs = m2 a(t) ≤ µs N = µs m2 g.474 (rad) = 27.8 m/s )× 2 (91/90)4 − 1 = 2.8% less than the angle itself.08 m/s . but ω1 = 2ω2 . A second block rests on the first.5). it is the frictional force which accelerates the upper block. A1 = 2A2 implies (a) ω1 = 1 ω2 and (b) a1. so this answer is only approximate. where Tmax = mg(1 + A2 ). a0 = (9. you suspend your keys from a thread and set the resulting pendulum oscillating.] Problem 66. In simple harmonic motion. so a1. A 500-g block on a frictionless surface is connected to a rather limp spring of constant k = 8.e. and A is the angular amplitude. so A < (6. If a0 is perpendicular to g (as for the air plane in this problem) then |g − a0 | = g 2 + a2 . What is the maximum allowable amplitude for pendulum motion of this system? figure 15-44 Problem 60. When the amplitude of the motion is increased to 35 cm.1◦. Two mass-spring systems have the same mass and the same total energy. Solution It is shown in Example 15-4 that the greatest tension in a simple pendulum occurs at the bottom of its swing. Since ω0 = g/ℓ at rest.44. The ratio of the frequencies is 2 ω/ω0 = 91/90. It completes exactly 90 cycles in 1 minute.1◦ = 0. then E1 = 4E2 .) Problem 62. The amplitude of system 1 is twice that of system 2. and now find the pendulum completes exactly 91 cycles in 1 minute. Find the plane’s acceleration. then ω1 A1 = ω2 A2 . b = 1.0 N. or 2 µs = (2π/1. A 500-g mass is suspended from a thread 45 cm long that can sustain a maximum tension of 6.CHAPTER 15 243 Solution The solution of Problem 21 shows that the period of a simple pendulum (point mass suspended in a constant gravitational field) in an accelerating reference frame is T = 2π/ω = 2π ℓ/ |g − a0 |. Solution (a) The energy of a mass-spring system is E = 1 2 2 2 mω A (see Section 15. The potential energy of a 75-g particle is given by U = ax2 − bx4 . and d2 x/dt2 = −(10ga/7)x. the child and swing rise slightly. However. is constant. (a) Show that there is a metastable equilibrium at x = 0. d2 U/dx2 = 2(a − 6bx2 ) is positive for x = 0. (c) What is the maximum amplitude for oscillatory (but not necessarily simple harmonic) motion to occur? Solution The potential energy of the ball is U = mgy = mgax2 . which is the equation for simple harmonic motion with frequency f = 2a/m/2π = 1. Hint: For small x. Solution The number of full turns (revolutions) is θ/2π. is constant.21 m. or dt2 Problem 66 Solution. Fx = −dU/dx = −2x(a − 2bx2 ) = md2 x/dt2 . in fact U → −∞ for |x| → ∞. Problem 68. Repeat Problem 46 for a small solid ball of mass M and radius R that rolls without slipping on the parabolic track. E = M gh + 1 Iω 2 = M g(1 cm)(θ/2π)2 + 2 1 2 2 I(dθ/dt) . (c) The motion is oscillatory as long as the particle is in the potential well surrounding x = 0 (see sketch). The total energy. These are x = 0 and x = ± a/2b = ±1. in general. (x = 0 is not an absolute minimum because U (x) < 0 for |x| > a/b. |x| < a/2b = 1. where θ is the angle (in radians) measured around the appropriate vertical axis. Problem 65 Solution. A child twirls around on a swing.) Note that in this problem (and in Problem 46) v = (dx/dt)2 + (dy/dt)2 .21 m. so the motion is not harmonic. and its kinetic energy (rolling without slipping) is 1 (from Equation 12-24) K = 1 M v 2 + 2 Icm ω 2 = 2 1 1 2 7 2 2 2 2 2 M v + 2 ( 5 M R )(v/R) = 10 M v . and x is in meters. representing unstable equilibria. so these points are relative maxima of potential energy. as shown in Fig. and the rotational inertia of the pair about the appropriate vertical axis is 0. x4 ≪ x2 . This is simple harmonic motion with period T = 2π 2π 2 I M g(1 cm) . so h = (1 cm)(θ/2π)2 . What is the period of this oscillation. v ≈ dx/dt. which deal with the negative of this potential energy function. it goes into torsional oscillation. dE d =0= dt dt 7 M v2 10 dx 7 dv = 2M gax + Mv . E = U + K. For small x. for small displacements.54 Hz. =− dt2 2π 2 I +I dθ dt d2 θ . however.244 CHAPTER 15 Problem 67. (See Problems 8-43 and 83. so dE θ dθ = 0 = M g(1 cm) 2 dt 2π dt d2 θ M g(1 cm) θ. 15-45.) (b) In onedimensional motion. As a result. which represents a relative minimum of the potential energy and therefore a point of metastable equilibrium. d2 U/dx2 < 0 for x = ± a/2b.55 J. h. When the child stops twisting up the swing. This is the case provided the total energy E < U (± a/2b) = a2 /4b = 2. and the maximum displacement from the origin. assuming that all the potential energy of the system is gravitational? The combined mass of the child and swing is 20 kg. twisting the swing ropes. which is simple harmonic motion with period T = 2π/ω = 2π 7/10ga. dt 5 dt M gax2 + Solution (a) The equilibrium points (for one-dimensional motion) are solutions of the equation dU/dx = 0 = 2x(a − 2bx2 ). (b) Find the frequency of small amplitude oscillations about this equilibrium.12 kg·m2 . 2bx2 ≪ a so d2 x/dt2 ≈ −(2a/m)x. with the rise. Since the total mechanical energy. in cm equal to the square of the number of full turns of the swing. we find φ = tan−1 [−v(tc )÷ ω2 x(tc )] − ω2 tc = tan−1 (68/3.7 ..7 m/s. Thus. Solving for φ (using sin / cos = tan). Just after the collision.) It follows from this that v(t) = −ω2 A× sin(ω2 t + φ). where g0 is the surface value.. The magnitude of the gravitational acceleration inside Earth is given approximately by g(r) = g0 (r/RE ). as asked in the problem.462 J. where just before the collision. where tc = π/2ω1 . x(tc ) = 10 cm (given) and v(tc ) = (m1 v1 + m2 v2 )/(m1 + m2 ). 1.2 kg) = 4. show that an object dropped into this hole executes simple harmonic motion. φ can also be found from either expression for x(tc ) or v(tc ). The amplitude A and phase constant φ can be determined from these two equations evaluated just after the collision.7 m/s)(0. Determine the frequency. we find A = x(tc )2 + [−v(tc )/ω2 ]2 = (10 cm)2 + (68 cm/3.2) = −68 cm/s.22◦ = −0. e. so an object in this hole (not subject to any .4 cm. E = K(tc ) + U (tc ) = 0. where ω2 and tc are known. and phase constant (relative to the original t = 0) of the resulting motion.1 m)2 = 1 0.6 (π/2) radians = 69. and RE is Earth’s radius. The collision is completely inelastic.8 + 1.39×22. since for the rightmost point of oscillation. ω2 √ tc = ω2 π/2ω1 = (π/2) ◦m1 /(m1 + m2 ) = 0. then conservation of momentum during the collision can be applied (see Equation 11-4). the acceleration is directed toward Earth’s center. x(t) = (10 cm) cos(ω1 t − π/2) = (10 cm) sin ω1 t. r is the distance from Earth’s center. i.2 kg 0.4).39 s−1 is the angular frequency when both blocks oscillate on the spring (and f2 = ω2 /2π = 0.91 s. 15-46). and v(tc ) = −68 cm/s = −ω2 A × sin(ω2 tc + φ). for t < tc . (This specifies the original zero of time appropriate to the given phase constant of −π/2.577 J)/(23 N/m). or ω1 tc = π/2. essentially at tc . if we assume that the collision takes place almost instantaneously. Solution The simple harmonic motion with just the first block on the spring can be described by Equation 15-9 and the given amplitude and phase constant.115 J. A 1. The block is oscillating with amplitude 10 cm and with phase constant φ = −π/2.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 23 N/m (Fig. sin ω1 tc = 1. where ω1 = k/m1 = (23 N/m)/(1.68 m/s)2 = 2 1 1 2 0. Once A is known. (Note: The solution for A is equivalent to calculating the various energies in the second simple harmonic motion. and the two blocks stick together.4) = sin−1 (68/3. This equation holds up to the time of the collision.e. amplitude. Evaluate and compare with the period of a satellite in a circular orbit not far above Earth’s surface. v(tc ) = (−1.8)/(0. x(t) = A cos(ω2 t + φ) for Solution The force of gravity is directed toward the center of the Earth. since just after the collision.01 m) 2 245 = 2π 2 = 6. 2 1 K(tc ) = 2 (m1 + m2 )v (tc ) = 1 (2 kg)(−0. the two equations become x(tc ) = 10 cm = A × cos(ω2 tc + φ).39×10) − 69.109 radians.) Solving for A (using sin2 + cos2 = 1). and find an expression for the period. where ω2 = k/(m1 + m2 ) = 3. A block of mass 0. Numerically.12 kg·m2 ) (20 kg)(9. or A = 2(0.) Equation 15-9 also describes the simple harmonic motion after the collision. ω2 tc + φ = cos−1 (10/22.) Problem figure 15-46 Problem 69.CHAPTER 15 2(0. (In fact. Suppose a narrow hole were drilled straight through the center of Earth and out the other side.577 J = 2 kA2 . figure 15-45 Problem 68 Solution. U (tc ) = 2 kx(tc ) = 2 (23 N/m)(0. Problem 69. It strikes the first block when the latter is at the rightmost point in its oscillation.39)2 = 22.7◦ = −6.7 m/s (also given).8 m/s )(0.g.7 m/s t > tc .80 kg is moving from the right at 1. 70. Neglecting air resistance.38 s−1 .80 kg 1.540 Hz. v1 = 0 (given m1 at rightmost point of its original motion) and v2 = −1. (This corresponds to the same initial conditions as for a simple pendulum.246 CHAPTER 15 T0 = 1.0042593 . For a pendulum with Solution Suppose that no forces with components in the direction of motion act on the mass other than the spring forces. Solution Numerical values are displayed in the table and sketch below. The · · · indicates that additional terms (in fact.0% .81 m/s ) = 84. By what percentage does the plotted period differ from T0 for 30◦ and 45◦ ? other forces) has an acceleration d2 r/dt2 = −(g0 /RE )r.000631 . 15-47. This is simple harmonic motion. Find the period of oscillations perpendicular to the springs. Since 2 2 3 g0 = GME /RE . At equilibrium. where s = Rθ. If m is given a small displacement .) The tangential component of Newton’s second law is −mg sin θ = m(d2 s/dt2 ). sin2 ( 1 θ 0 ) 2 . L m L Problem 72.4 min . This represents one-dimensional simple harmonic motion about r = 0 (compare to Equation 15-4). T = 2π RE /GME . so ω = g/R and T = 2π R/g. A small object of mass m slides without friction in a circular bowl of radius R.4% 1. with the same equations as for a simple pendulum (see Problem 24). and compare with that of a simple pendulum. For small angles.37×106 m)/(9.016747 . with period T = 2π RE /g0 = 2π (6. and θ0 is the amplitude. Problem 73. assuming sufficiently small amplitude that the magnitude of the spring tension is essentially unchanged. the same as for a satellite in an Earth-grazing orbit. where T0 = 2π ℓ/g is the period in the limit of arbitrarily small amplitude.0000408 . infinitely many more) are needed for an exact expression. −g sin θ ≈ −gθ ≈ R(d2 θ/dt2 ).7% 4. θ0 15 30◦ 45◦ ◦ 1 4 Problem 71. as shown in Fig. the tension force in each spring is F0 . Solution What is intended in this problem is consideration of just motion in a vertical plane through the axis of the bowl. Problem 71 Solution.03661 9 64 1 sin4 ( 2 θ0 ) (T − T 0 )/T 0 0. plot the period given above versus amplitude for amplitudes from 0 to 45◦ . A more exact expression than Equation 15-18 for the period of a simple pendulum is T = T0 1 + 1 sin2 4 1 θ0 2 + 9 sin4 64 1 θ0 2 + ··· . Derive an expression for small-amplitude oscillations about equilibrium.00 s. A mass m is connected between two springs of length L.00302 Problem 72 Solution. Other conditions lead to motion like various conical pendulums. figure 15-47 Problem 73. − g 2gℓ2 1 2 2MR + M ℓ2 M gℓ This is a minimum when d dℓ =0= √ or ℓ = R/ 2. or by noting that T → ∞ for ℓ → 0 and ℓ → ∞. The wire is placed over a pivot. Solution The period of the physical pendulum shown is T = 2π = 2π I = 2π M gℓ ℓ R2 + .CHAPTER 15 perpendicular to the springs (as sketched). the net force is Fy = −2F0 sin θ = −2F0 y/ L2 + y 2 ≈ −2F0 y/L. The rotational inertia of the bent wire (two thin rods) about the pivot is I = 1 1 2[ 3 ( 1 M )ℓ2 ] = 3 M ℓ2 . 15-48). and so has a minimum between. Equation 15-16 for the physical pendulum gives ω= M gh = I M g(ℓ/2) cos(θ/2) = M ℓ2 /3 3g cos(θ/2) 2ℓ Problem 73 Solution. 2ℓ Solution The CM of the bent wire is a distance h = (ℓ/2)× cos(θ/2) from the pivot. 247 Problem 75. 15-49. for y ≪ L.) . A disk of radius R is suspended from a pivot somewhere between its center and edge (Fig. Newton’s second law gives md2 y/dt2 = Fy . or d2 y/dt2 ≈ −(2F0 /mL)y. This is the equation for simple harmonic motion with angular frequency ω = 2F0 /mL and period T = 2π/ω = 2π mL/2F0 . For what pivot point will the period of this physical pendulum be a minimum? Suspension point ? figure 15-48 Problem 74 Solution. Problem 74. g 2gℓ ℓ R2 + g 2gℓ 1 R2 . where M is the mass of the whole 2 wire. as shown in Fig. Show that the angular frequency of smallamplitude oscillations about this equilibrium is given by ω= 3g cos(θ/2) . (The fact that this represents a minimum can be seen either by calculating d2 T ÷ dℓ2 > 0. A uniform piece of wire is bent into a V-shape with angle θ between two legs of length ℓ. and show that your answer is just the maximum velocity. Solution The time of the maximum (positive) displacement is given by the equation ωt1 + φ = 0. Integrating the nonconstant acceleration of a harmonic oscillator over time from the time of maximum displacement to the time of zero displacement should give the velocity at zero displacement. Thus. t2 t2 a dt = t1 t1 (−ω 2 A) cos(ωt + φ)dt t2 t1 = − ω2A sin(ωt + φ) ω = −ωA sin(π/2) + ωA sin(0) = −ωA. and the time of the next zero displacement by ωt2 + φ = π/2. using Equation 15-11 for the acceleration. Carry out this integration. .248 CHAPTER 15 figure 15-49 Problem 75 Solution. it is negative because we assumed the oscillator started at +A. which occurs at zero displacement. where the displacement is given by Equation 15-9. Problem 76. This is the maximum velocity.