Homework 4 (Chapter 15) Homework 4 (Chapter 15) Wave and Particle Velocity Vector Drawing A long string is stretched and its left end is oscillated upward and downward. Two points on the string are labeled A and B. Part A ⃗ and v ⃗ , to correctly represent the Points A and B are indicated on the string. Orient the two vectors, v A B direction of the wave velocity at points A and B. Rotate the given vectors to indicate the direction of the wave velocity at the indicated points. Hint 1. Distinguishing between wave velocity and particle velocity A wave is a collective disturbance that, typically, travels through some medium, in this case along a string. The velocity of the individual particles of the medium are quite distinct from the velocity of the wave as it passes through the medium. In fact, in a transverse wave such as a wave on a string, the wave velocity and particle velocities are perpendicular. Hint 2. Wave velocity A wave on a stretched string travels away from the source of the wave along the length of the string. ANSWER: Correct 1/33 Homework 4 (Chapter 15) Part B ⃗ and v ⃗ to correctly represent the direction of the velocity of At the instant shown, orient the given vectors v A B points A and B. At each of the points A and B, rotate the given vector to indicate the direction of the velocity. Hint 1. Distinguishing between wave velocity and particle velocity A wave is a collective disturbance that, typically, travels through some medium, in this case along a string. The velocity of the individual particles of the medium are quite distinct from the velocity of the wave as it passes through the medium. In fact, in a transverse wave such as a wave on a string, the wave velocity and particle velocities are perpendicular. Hint 2. Determining velocity from a snapshot The diagram represents the position of a small portion of the string at a specific instant of time: a snapshot of the string at this time. Based on only a snapshot, you cannot determine the velocity of an object, such as a point on the string. However, you also know that the left end of the string is the source of the wave disturbance. From this information you can deduce what is about to happen to point A’s position, and from this change in position deduce the direction of point A’s velocity (and similarly for point B). Hint 3. Find the change in point A’s position Based on the location of the source of the wave (the left end of the string), will the wave crest to the immediate left of point A soon raise or lower point A’s position? ANSWER: raise lower Hint 4. Find the change in point B’s position Based on the location of the source of the wave (the left end of the string), will the wave trough to the immediate left of point B soon raise or lower point B’s position? ANSWER: raise lower ANSWER: 2/33 Homework 4 (Chapter 15) Correct Exercise 15.4 Ultrasound is the name given to frequencies above the human range of hearing, which is about 20000 Hz . Waves above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel with a speed of 1500 m/s. For a good detailed image, the wavelength should be no more than 1.0 mm. Part A What frequency is required? ANSWER: f = 1.50×106 Hz Correct ± Speed of Rowing and Swimming A displacement boat (generally a moving object held up by buoyancy of the water) generates a wave that travels alongside at the speed of the boat with the first wave crest at the bow of the boat. Because the wavelength increases with the boat's speed, above a certain speed the second crest appears astern of the boat, so that the boat must travel uphill on the wave it creates. If it lacks the power to get over the top of this wave, its speed is limited to roughly the speed of a wave with the length of the boat. Hence we have the following rule of thumb: The maximum speed of a modestly powered boat is approximately equal to (actually, a little faster than) the wave 3/33 the propagation velocity of water waves is − − v= √ g k . where k is the wavenumber and g is the gravitational constant. to three significant figures. Hint 1. in seconds. Estimate the speed of the boat Find v 8 . the maximum speed of the eight. Empirically. Hint 1. Express your answer numerically.81m/s .6m long with eight rowers and a coxswain who steers and calls the stroke. Express the velocity in terms of λ and constants like g and π. Express the time numerically. Part A Find the velocity of water waves v in terms of the wavelength λ . Find k in terms of λ Express the wavenumber k in terms of λ .24 m/s 4/33 . Your answer may also include constants like π. Part B Estimate the time T 8 it takes an "eight" to travel 2000 m. ANSWER: k 2π = λ ANSWER: −− − v = √ gλ 2π Correct An eightperson shell (or "eight") is a light rowing craft approximately 17.Homework 4 (Chapter 15) whose wavelength equals the length of the boat measured at the waterline. in meters/second. to three significant figures. Assume that the speed of the shell is limited according to the rule given in the introduction. use 2 g = 9. ANSWER: v8 = 5. In parts that ask for numerical answers. Estimate the speed of the swimmer Find the maximum speed v 100 of the swimmer. to swim 100 m. in seconds. One formula for a wave with a y displacement (e.74 s for the freestyle and 57.Homework 4 (Chapter 15) ANSWER: T8 = 382 s Correct A good time for an eight over this distance (one typically used in US college crew races) is 390 s.7 s Correct As of 2002.80m tall. They give estimates valid at the 10% or 20% level.47 s for the breaststroke. of a string) traveling in the x direction is y(x. accurate to three significant figures.. according to the rule given in the problem introduction. Hint 1. 1. to three significant figures. All the questions in this problem refer to this formula and to the wave it describes.g. the world record for a swimmer who kicks off the walls of a 25m pool four times is 46. Part C Estimate the time T 100 it takes a good human swimmer.68 m/s ANSWER: T 100 = 59. t) = A sin(kx − ωt) . Assume that the speed of the swimmer is limited by his or her height. These examples show the power of simple rules. Standard Expression for a Traveling Wave Learning Goal: To understand the standard formula for a sinusoidal traveling wave. in meters/second. Express the time numerically. Part A Which of the following are independent variables? 5/33 . but these estimates apply over a wide range of situations that are fundamentally the same (from the point of view of the underlying physics). ANSWER: v 100 = 1. Express your answer numerically. x .Homework 4 (Chapter 15) Hint 1. and period of the wave. frequency. k . ANSWER: x t only only A k only ω x only only and t ω and t A and k and ω Correct Part C What is the phase ϕ(x. ANSWER: x t only only A k only ω x only only and t ω and t A and k and ω Correct Part B Which of the following are parameters that determine the characteristics of the wave? Hint 1. t) of the wave? Express the phase in terms of one or more given variables ( A. For a wave these include the amplitude. What are independent variables? Independent variables are those that are freely varied to control the value of the function. t . wavelength. What are parameters? Parameters are constants that determine the characteristics of a particular function. and ω ) and any needed 6/33 . The independent variables typically appear on the horizontal axis of a plot of the function. sloping upward at x = 0.e. k . the length of one complete cycle of oscillation). and ω) and any needed constants like π. Hint 1. t. t. The wavelength is the x position at which the wave next crosses the y axis. Hint 1. x. ANSWER: λ = 2π k Correct Part E What is the period T of this wave? Express the period in terms of one or more given variables ( A. k . which is expressed in radians. sloping upward (i. t) = kx − ωt Correct Part D What is the wavelength λ of the wave? Express the wavelength in terms of one or more given variables ( A. k . t . Definition of phase The phase is the argument of the trig function.Homework 4 (Chapter 15) constants like π. x. x . and ω) and any needed constants like π. ANSWER: T = 2π ω Correct Part F What is the speed of propagation v of this wave? Express the speed of propagation in terms of one or more given variables ( A. and ω ) and any 7/33 . ANSWER: ϕ(x.. The wave crosses the y axis. Finding the wavelength Consider the form of the wave at time t = 0 . a tightrope walker moves her foot up and down near the end of the tightrope. when will the ant become "weightless"? ANSWER: When it has no net force acting on it When the normal force of the string equals its weight When the normal force of the string equals twice its weight When the string has a downward acceleration of magnitude g 8/33 . Think about what happens to the normal force in this situation.Homework 4 (Chapter 15) needed constants like π. How to approach the problem In the context of this problem. Weight and weightless Weight is generally defined as being the equal and opposite force to the normal force. Note that the force due to gravity does not change and would still be the same as when the elevator was static. ANSWER: v = ω k Correct Ant on a Tightrope A large ant is standing on the middle of a circus tightrope that is stretched with tension T s . in a falling elevator) it would read zero. μ .g. generating a sinusoidal transverse wave of wavelength λ and amplitude A. On a flat surface in a static situation. "Weightless" is a more colloquial term meaning that if you stepped on a scale (e. λ . Part A What is the minimum wave amplitude Amin such that the ant will become momentarily "weightless" at some point as the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wave propagation. Hint 1. Hint 2. Express the minimum wave amplitude in terms of T s . you have found the wavelength and the period of this wave. Wanting to shake the ant off the rope. Hint 1.. The speed of propagation is a function of these two quantities: v = λ/T . and g . Assume that the magnitude of the acceleration due to gravity is g . How to find v If you've done the previous parts of this problem. the weight is equal to the force due to gravity acting on a mass. The rope has mass per unit length μ . To solve this problem. k . to a lesser degree.Homework 4 (Chapter 15) Correct The ant will become "weightless" when the normal force between the string and the ant becomes zero. When an elevator accelerates downward. Once this happens. (At some time. The first derivative Differentiate the given equation for the displacement of the string y(x. t. Hint 1. Hint 2. you must determine the amplitude for which the maximum acceleration of a point on the string is equal to −g . since the normal force on us from the elevator floor goes to zero. gravity is the only force acting on the ant; the ant will become "weightless" and will accelerate downward under the influence of gravity alone. t) acceleration a max of a point on the string? = A sin(ωt − kx) . t) to find the vertical velocity v y (x. Hint 1. Find the maximum acceleration of the string Assume that the wave propagates as y(x. twice. Then determine what the maximum value of this acceleration is. t) of the rope. How to approach the problem Use the formula given for the displacement of the string to find the acceleration of the string as a function of position and time. and x. ω. and x. t) = ωAcos(ωt − kx) ANSWER: a y (x. Hint 3. the bit of rope underneath the ant will have this maximum downward acceleration. we feel lighter. Express your answer in terms of A. in an elevator. ω. k . Acceleration of a point on the string Find the vertical acceleration a y (x. t. If this downward acceleration is equal to g then we feel effectively "weightless". t). t) of an arbitrary point on the string as a function of time.) Hint 2. Express your answer in terms of A. How to find the acceleration Differentiate the expression given for y(x. the displacement of a point on the string. t) = 2 ω Asin(kx − ωt) 9/33 . What is the maximum downward Express the maximum downward acceleration in terms of π and any quantities given in the problem introduction. This same effect can be observed. ANSWER: v y (x. Hint 2. and velocity. Hint 4. Find the maximum downward acceleration The maximum downward acceleration a max is the most negative possible value of a y (x. and a formula for the velocity of a wave on a string to find an expression for ω in terms of quantities given in the problem introduction. a relationship among frequency.Homework 4 (Chapter 15) Correct Hint 3. wavelength. μ . λ . As the wave passes beneath the ant. Relationship among frequency. and velocity of a wave are related by v = λf . wavelength. ANSWER: a max = 2 −ω A Hint 4. t) . and π. Express the angular frequency in terms of T s . find an expression for ω in terms of given quantities. To solve the problem. When will the acceleration reach its most negative value? The most negative acceleration occurs when sin(ωt − kx) = 1 . wavelength. Hint 1. Hint 3. General formula for ω The angular frequency of a wave is equal to 2π times the normal frequency: ω = 2πf . How to approach this question Combine a general formula for ω. and velocity The frequency. Hint 1. Speed of a wave on a string What is the speed v of any wave on the string described in the problem introduction? ANSWER: −− − v = √ Ts μ ANSWER: 10/33 . What is a max ? Express your answer in terms of ω and quantities given in the problem introduction. Determine ω in terms of given quantities The angular frequency ω of the wave in the string was not given in the problem introduction. at some time or another the ant will be at a point where the acceleration of the string has this most negative value. and wavelength λ = 0.320 m . amplitude A = 0. ANSWER: 11/33 .0700 m . Putting it all together Once you have an expression for the maximum acceleration of a point on the string a max . ANSWER: 2 A min = μg( λ ) 2π Ts Correct Exercise 15. determine what amplitude is required such that a max = −g.7 Transverse waves on a string have wave speed v = 8. This will be the minimum amplitude Amin for which the ant becomes weightless. and at t = 0 thex = 0 end of the string has its maximum upward displacement.Homework 4 (Chapter 15) −− − ω = √ Ts 2π μ λ ANSWER: a max = − Ts μ 2 ( 2π λ ) A Incorrect; Try Again; 9 attempts remaining Hint 4.00 m/s. ANSWER: f = 25. Part A Find the frequency of these waves.0 Hz Correct Part B Find the period of these waves. The waves travel in the x direction. 00×10−2 s Correct Part C Find the wave number of these waves.6x + 157t) m Correct Part E Find the transverse displacement of a particle at x = 0.360 m next has maximum Express your answer using three significant figures.00×10−3 s 12/33 .Homework 4 (Chapter 15) 4. ANSWER: t = 5. Express your answer in terms of the variables x and t.07cos(19. ANSWER: k = 19.360 m at time t = 0. t) = 0.150 s .6 rad/m Correct Part D Write a wave function describing the wave. ANSWER: y(x. ANSWER: y = 4.95×10−2 m Correct Part F How much time must elapse from the instant in part E until the particle at x upward displacement? = 0. and the string is stretched by a force of magnitude T s applied to the first segment; T s is much greater than the total weight of the string. made up of three segments of equal length. Hint 1. and any constants. Example: speed in the second segment Find v 2 . 13/33 . Which quantities are the same and which are different in each of the three segments? Hint 1. General formula for the speed of a transverse wave on a string Consider a section of string with mass per unit length μ under tension T s . The third segment is tied to a wall. How do the segments differ? Consider each segment of the string seperately. and T s . General formula for the speed of a transverse wave on a string Consider a section of string with mass per unit length μ under tension T s . The speed v with which a transverse wave travels along this segment of the string is given by −− − v= √ Ts μ . Express the wave speed on the second segment in terms of T s . The speed v with which a transverse wave travels along this segment of the string is given by −− − v= √ Ts μ . that of the second is 2μ . and that of the third μ/4 . The mass per unit length of the first segment is μ . Both the wave velocity and the tension are different in the three segments. ANSWER: Both the wave velocity and the tension in the string are the same in all three segments. Part A How long will it take a transverse wave to propagate from one end of the string to the other? Express the time t in terms of L. Hint 2. μ . Hint 1. the speed of propagation of the wave in the second segment of the string described in the problem introduction. The tension is the same in each segment but the wave velocity is different. μ . The wave velocity is the same in each segment but the tension is different.Homework 4 (Chapter 15) Correct Wave Propagation in a String of Varying Density Consider a string of total length L . and the wave will experience them as boundaries. Express the velocity of propagation in terms of some or all of the variables A. is equal to zero at (x.Homework 4 (Chapter 15) ANSWER: −− − v2 = √ Ts 2μ Hint 3. Some math help Recall that a b/c = a⋅c b and − − − − − −−−− √m + √n ≠ √m + n . Although this will not affect the time it takes for the wave to reach the end of the string (thus it is not directly relevant to this question). Also. after the main wave has arrived. air or a string) and the velocity of the medium (the air or the string itself). we may observe later arrivals of waves that have reflected back and forth between the boundaries before finally reaching the end of the string. ω. Consider a transverse wave traveling in a string.. The mathematical form of the wave is y(x. amplitude) to be reflected. Part A Find the speed of propagation v p of this wave. the wave's amplitude will be reduced. while the rest is transmitted at each boundary. Hint 1. Perform an intermediate step Note that the phase of the wave (kx − ωt). t) = (0. k .g. t) = A sin(kx − ωt) . 0) . At what position x = Δx is the phase equal to zero a short time t = Δt later? Express your answer in terms of Δt. and ω. ANSWER: 14/33 . Two Velocities in a Traveling Wave Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e. ANSWER: t = √ L L 3 3 + Ts μ √ Ts 2μ L 3 + √ 4T s μ Correct The changes in density along the string are sudden. This will cause a fraction of the wave (energy. and k . and therefore the displacement of the string. take the partial derivative of y(x. Express the y velocity in terms of ω. A helpful derivative d dt sin(at + b) = a cos(at + b) ANSWER: v y (x. t) of a point on the string as a function of x and t. t). t). ANSWER: 15/33 . take the time derivative of y(x. You are not given any information about Δx(x. x. and t. t) with respect to time. t) . How to approach this question You are given a form for y(x. is true? Hint 1. you are given an expression for y(x. How to approach the problem In the problem introduction. the displacement of the string as a function of x and t. the x component of the velocity of the string. To find the y velocity. That is. t). t) while treating x as a constant.Homework 4 (Chapter 15) Δx = ωΔt k ANSWER: vp = ω k Correct Part B Find the y velocity v y (x. but it is assumed that each point on the string only moves in the y direction. Δx(x. k . Hint 1. t) = 0. t) = −ωAcos(kx − ωt) Correct Part C Which of the following statements about v x (x. i. A. Hint 2.e. t) = − ω k ∂x 16/33 . and v p . and t.Homework 4 (Chapter 15) v x (x. t) but is 180 ∘ out of phase. v x (x. A helpful derivative d dx cos(ax + b) = −a sin(ax + b) ANSWER: ∂y(x.t) ∂y(x. t) = v y (x. a given shape or pattern of points (in this sinusoidal) will move to the right as time progresses. t) = v p v x (x.t) ∂x as a function of position x and time t. t) = 0 Correct So the wave moves in the x direction. What this means is that even though individual points on the string only move up and down. Hint 1. Express your answer as a suitable combination of some of the variables ω.t) ∂x = Akcos(kx − ωt) Correct Part E Find the ratio of the y velocity of the string to the slope of the string calculated in the previous part. t) v x (x. even though the string does not. ω. Part D Find the slope of the string ∂y(x. t) has the same mathematical form as v y (x. ANSWER: v y (x. x. k . Express your answer in terms of A. k . Homework 4 (Chapter 15) Correct To understand why the ratio of the y velocity of the string to its slope is constant. ANSWER: 17/33 . the string in the vicinity of x = 0 will be sloped downward. If the slope at (x.6 m away. t) < 0 ).25 μW/m2 Correct Part C What power of sound does the jet produce at takeoff? Express your answer using two significant figures. that bit of string will be moving upward. Assume that the plane behaves like a point source of sound. t)/∂x > 0). the string is sloped upward. In general. that bit of string will be moving downward (v y (x. One way of understanding why the ratio has a constant magnitude is to observe that the more steeply the string is sloped. ANSWER: I = 0. Exercise 15. In the vicinity of x = 0. if at some particular (x.0 μW/m . Part A What is the closest distance you should live from the airport runway to preserve your peace of mind? Express your answer using two significant figures. which is 1. and the string at position x = 0 will move upward as the wave moves forward. the more quickly it will move up or down. t) is negative.25 A jet plane at takeoff can produce sound of intensity 10. This explains why the sign of the ratio of string velocity to slope is always negative.0 W/m2 at 30. draw the string with a wave running along it at time t = 0 . ANSWER: r = 97 km Correct Part B What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? Express your answer using two significant figures. Onehalf cycle later. The bit of string at position x = 0 moves downward as the wave moves forward. But you prefer the tranquil 2 sound of normal conversation. t) the slope of the string is positive (∂y(x. 34 Two pulses are moving in opposite directions at 1.0 cm/s on a taut string.22 A piano wire with mass 2.2×105 W Correct Exercise 15.204 W Correct Part B What happens to the average power if the wave amplitude is halved? ANSWER: P = 5. Each square is 1. ANSWER: P = 0.0 N .11×10−2 W Correct Exercise 15.Homework 4 (Chapter 15) P = 1. Part A Calculate the average power carried by the wave.0 cm is stretched with a tension of 32.55 g and length 80. 18/33 . as shown in the figure .0 cm.50 mm travels along the wire. A wave with frequency 120 Hz and amplitude 1. Homework 4 (Chapter 15) Part A Sketch the shape of the string at the end of 6.0 s ; ANSWER: 19/33 .0 s ; ANSWER: Part B Sketch the shape of the string at the end of 7. Homework 4 (Chapter 15) Part C Sketch the shape of the string at the end of 8.0 s . ANSWER: 20/33 . The wave is traveling but not oscillating.Homework 4 (Chapter 15) Correct Creating a Standing Wave Learning Goal: To see how two traveling waves of the same frequency create a standing wave. Consider a traveling wave described by the formula y (x. where y1 (x. a local electric field. Consider the sum of two waves y1 (x. t) . The wave is traveling in the −x direction. the wave described in the problem introduction. then the sum (or difference) also satisfies the wave equation. t) + y2 (x. the position of the surface of a body of water. Part A Which one of the following statements about the wave described in the problem introduction is correct? ANSWER: The wave is traveling in the +x direction. The wave is oscillating but not traveling. t) . These waves have been chosen so that their sum can be written as follows: (x. Correct Part B Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling in the opposite direction? At time t = 0 this new wave should have the same displacement as y1 (x. This function might represent the lateral displacement of a string. This principle follows from the fact that every term in the wave equation is linear in the amplitude of the wave. t) = A sin(kx − ωt) 1 . ANSWER: A cos(kx − ωt) A cos(kx + ωt) A sin(kx − ωt) A sin(kx + ωt) Correct The principle of superposition states that if two functions each separately satisfy the wave equation. t) is the wave described in Part B. or any of a number of other physical manifestations of waves. t) = (x) (t) 21/33 . t) is the wave described in Part A and y2 (x. what is the displacement of the string (assuming that the standing wave ys (x. Applying the identity Since you really need an identity for sin(A − B). Hint 1. x. keeping in mind that sin(−x) = − sin(x). t) is correct? ANSWER: This wave is traveling in the +x direction. This form is significant because ye (x) . Part C Find ye (x) and yt (t). depends only on position. k . it oscillates up and down only. Express your answers in terms of A. This wave is traveling in the −x direction. simply replace B by −B in the identity from Hint C. t) is Express your answer in terms of parameters given in the problem introduction. Traditionally. yt (t) = 2Asin(kx). A useful identity A useful trigonometric identity for this problem is sin(A + B) = sin(A) cos(B) + cos(A) sin(B) . and yt (t) depends only on time. Part E At the position x present)? = 0 . the wave does not appear to move left or right; rather. t) = y (x)y (t) s e t .Homework 4 (Chapter 15) y (x. Separate the two functions with a comma. This wave is oscillating but not traveling. Because each part of the string oscillates with the same phase. called the envelope. Hint 2. ANSWER: y (x) e . the overall amplitude of the wave is written as part of ye (x) .1. Correct A wave that oscillates in place is called a standing wave. ANSWER: 22/33 . ω. This wave is traveling but not oscillating. cos(ωt) Correct Part D Which one of the following statements about the superposition wave ys (x. Keep in mind that yt (t) should be a trigonometric function of unit amplitude. the time function is taken to be a trigonometric function with unit amplitude; that is. and t. Which of the following statements does the string's being straight imply about the energy stored in the string? a. Part F At certain times. For any other value of cos(ωt) . causing the standing wave to form at a later time. ANSWER: a b c d Correct ± Why the Highest Piano Notes Have Short Strings 23/33 . ANSWER: t1 = π 2ω Correct Part G From Part F we know that the string is perfectly straight at time t = π 2ω . the string will be perfectly straight. ys (x. Hint 1. d. k . π c. 2ω Therefore.Homework 4 (Chapter 15) y s (x = 0. Such solutions will be important in treating normal modes that arise when there are two such constraints. since it could represent a string tied to a post or otherwise constrained at position x = 0. There is no energy stored in the string: The string will remain straight for all subsequent times. Express t1 in terms of ω. there is energy stored in the string. How to approach the problem The string can be straight only when cos(ωt) = 0 . The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle. t) will be a sinusoidal function of position x. Find the first time t1 > 0 when this is true. b. t) = 0 Correct This could be a useful property of this standing wave. Although the string is straight at time t = . parts of the string have nonzero velocity. for then ys (x. t) = 0 also (for all x). Energy will flow into the string. and necessary constants. ANSWER: v = −− − √ pT ρ ANSWER: 24/33 . ANSWER: − − − v = √ FT μ Correct Hint 2. N/m 2 and a density ρ of Part A What is the speed v of a wave traveling down such a wire if the wire is stretched to its breaking point? Express the speed of the wave numerically. in meters per second. There is only one combination of FT and μ that has the correct dimensions. Find the traveling wave speed in a stretched string What is the speed v of a traveling wave in a stretched string which has mass per unit length μ and is under tension FT ? Answer in terms of FT and μ .Homework 4 (Chapter 15) The steel used for piano wire has a breaking (tensile) strength pT of about 3 × 10 9 3 7800 kg/m . Hint 1. Note that this method does not give the value of any constant multiplier of the relation. Use dimensional analysis One way to answer this type of problem is to consider the dimensions of the quantities involved. The tensile strength pT and tension in the wire at the breaking point FT are related by pT A = FT . whereas we require an answer with dimensions of m ⋅ s −1 . The density ρ and mass per unit length μ are related by ρA = μ . Express v in terms of tensile strength and density What is the speed v of a traveling wave in terms of the tensile strength pT and wire density ρ ? Hint 1. Hint 1. How to approach the problem Assume that the wire has crosssectional area A. to the nearest integer. Force and mass per unit length have dimensions kg ⋅ m ⋅ s −2 and kg ⋅ m−1 respectively. Use these expressions to convert the speed in terms of FT and μ into a speed in terms of pT and ρ . you found the speed of traveling waves in the wire ( v ). A length constraint given f and v In Part A of this problem. You also know the frequency of oscillation ( f ). ANSWER: λ v = f Hint 2.75 cm Correct Normal Modes and Resonance Frequencies Learning Goal: 25/33 . what must the vibrating length of the wire be? Express the length numerically. Relationship between wavelength and string length Consider the boundary conditions for a stretched piano wire: Both ends are fixed. ANSWER: λ = 2L ANSWER: L = 7. If the wire is in tune when stretched to its breaking point. using three significant figures. in terms of these quantities? Express your answer in terms of v and f . unlike steel. If such a wire is oscillating at its fundamental frequency (its first normal mode). in piano wire the Part B Imagine that the wire described in the problem introduction is used for the highest C on a piano ( C8 ≈ 4000 Hz ). in centimeters. What is the wavelength ( λ ) of the wave in the C8 piano string. the wavelength will not be equal to the wire length. m/s ) because. Hint 1. What is the wavelength of the first normal mode of a string of length L that is fixed at both ends? Express the wavelength in terms of L.Homework 4 (Chapter 15) v = 620 m/s Correct This is much less than the speed of sound in steel (5941 tensile strength is much less than the Young's modulus. The wave is traveling in the x direction. t) and yi (L.Homework 4 (Chapter 15) To understand the concept of normal modes of oscillation and to derive some properties of normal modes of waves on a string. t) = 0. which correspond to the string being fixed at its two ends. The wave will satisfy the given boundary conditions for any arbitrary wavelength λ i . Consider an example of a system with normal modes: a string of length L held fixed at both ends. the time dependance of a normal mode is always sinusoidal. but the spatial dependence need not be. and the waves are transverse with displacement along the y direction. The string extends in the x direction. In this problem. yi (0. Ai must be chosen so that the wave fits exactly on the string. The wave does not satisfy the boundary condition yi (0. t) = A i sin(2π x λi ) sin(2πf i t). Any one of Ai or λ i or f i can be chosen to make the solution a normal mode.) Specifically. t) = 0 . Correct Part B Which of the following statements are true? ANSWER: The system can resonate at only certain resonance frequencies f i and the wavelength λ i must be such that yi (0. In general there are an infinite number of such modes. you will investigate the shape of the normal modes and then their frequency. Normal mode constraints The key constraint with normal modes is that there are two spatial boundary conditions. located at x = 0 and x = L. t) = yi (L. 26/33 . (For linear systems. Which of the following statements about the wave in the string is correct? Hint 1. each one with a distinctive frequency f i and associated pattern of oscillation. The wavelength λ i can have only certain specific values if the boundary conditions are to be satisfied. Assume that waves on this string propagate with speed v . The normal modes of this system are products of trigonometric functions. for this system a normal mode is described by y i (x. = 0 ANSWER: The wave is traveling in the +x direction. Part A The string described in the problem introduction is oscillating in one of its normal modes. A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. t) = 0. 3 Correct Hint 3. 27/33 . nonzero values of z that satisfy the equation sin(z) = 0 . Values of z that satisfy sin(z) = 0 The spatial part of the normal mode solution is a sine wave. z 2 . z 3 = 1. The three largest wavelengths that satisfy this equation at the other end of the string (with x = L) are given by 2πL/λ i = z i . Picture of the normal modes Consider the lowest four modes of the string as shown in the figure.2. those that satisfy the boundary conditions at x = 0 and x = L. z 2 .Homework 4 (Chapter 15) Correct The key factor producing the normal modes is that there are two spatial boundary conditions. List them in increasing order. Express the three nonzero values of z as multiples of π. and λ 3 ) that "fit" on the string. How to approach the problem The nodes of the wave occur where sin(2π x λi ) = 0. that are satisfied only for particular values of λ i . ANSWER: z1 . List them in decreasing order of length. These longest wavelengths have the lowest frequencies. since sin(0) = 0 . yi (0. t) = 0. and z 3 ) that satisfy sin(z) = 0 . Find the three smallest (nonzero) values of z (call them z 1 . Express the three wavelengths in terms of L. λ 2 . Hint 2. This equation is trivially satisfied at one end of the string (with x = 0 ). Hint 1. separated by commas. = 0 Part C Find the three longest wavelengths (call them λ 1 . where the z i are the three smallest. t) and yi (L. that is. separated by commas. 2L 3 Correct The procedure described here contains the same mathematics that leads to quantization in quantum mechanics. Nevertheless. L. ANSWER: f i = v λi 28/33 . which is characterized by its wavelength λ i . λ 3 = 2L. Propagation speed for standing waves Your expression will involve v .) The letter A is written over the antinodes. Hint 1. Find the frequency f i of the ith normal mode. ANSWER: λ1 . the speed of wave propagation is a physical property that has a welldefined value that happens to appear in the relationship between frequency and wavelength of normal modes. which are where the oscillation amplitude is maximum. Use what you know about traveling waves The relationship between the wavelength and the frequency for standing waves is the same as that for traveling waves and involves the speed of propagation v . the speed of propagation of a wave on the string. λ 2 . (Note that there are nodes in addition to those at the end of the string.Homework 4 (Chapter 15) The letter N is written over each of the nodes defined as places where the string does not move. Part D The frequency of each normal mode depends on the spatial part of the wave function. Hint 2. Express f i in terms of its particular wavelength λ i and the speed of propagation of the wave v . Of course. the normal modes are standing waves and do not travel along the string the way that traveling waves do. f 2 .400 m long and has a mass of 3. Part A What is the frequency of its fundamental mode of vibration? ANSWER: f = 408 Hz Correct Part B What is the nurmber of the highest hamonic that could be heard by a person who is capable of hearing frequencies up to 10000 Hz? ANSWER: 29/33 .00 g. f 3 = v 2L . and they are called partials. the highest audible normal frequencies for a given string can be a significant fraction of a semitone sharper than a simple integer multiple of the fundamental. t) . Part E Find the three lowest normal mode frequencies f 1 .40 A piano tuner stretches a steel piano wire with a tension of 800 N. ANSWER: f 1 .Homework 4 (Chapter 15) Correct The frequencies f i are the only frequencies at which the system can oscillate. these frequencies are multiples of the lowest frequency. which are equal to the energy of that mode divided by Planck's constant −34 h. Exercise 15. the fundamental frequencies of the lower notes are deliberately tuned a bit flat so that their higher partials are closer in frequency to the higher notes. Consequently. In SI units. that is. Express the frequencies in terms of L. For this reason the lowest frequency is called the fundamental and the higher frequencies are called harmonics of the fundamental. List them in increasing order. If the string is excited at one of these resonance frequencies it will respond by oscillating in the pattern given by yi (x. separated by commas. In quantum mechanics these frequencies are called the eigenfrequencies. In an acoustic piano. v . the stiffness of the string) are not valid. Planck's constant has the value h = 6. and any constants. The steel wire is 0. and f 3 . the normal mode frequencies are not exactly harmonic.63 × 10 J ⋅ s. 3v 2L Correct Note that. for the string. f 2 . v L . with wavelength λ i associated with the f i at which it is excited. When other physical approximations (for example. Homework 4 (Chapter 15) N = 24 Correct Exercise 15.0340 rad/cm)x] sin[(50. taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x.60 cm) sin[(0.41 A thin. ANSWER: λ = 185 cm Correct Part D Find the frequency of the traveling waves.80 cm Correct Part B What is the length of the string? ANSWER: L = 277 cm Correct Part C Find the wavelength of the traveling waves. ANSWER: A = 2. t) = (5. the xaxis is along the string and the yaxis is perpendicular to the string. Part A Find the amplitude of the two traveling waves that make up this standing wave. ANSWER: 30/33 . where the origin is at the left end of the string.0 rad/s)t]. launch the video below. 31/33 .Homework 4 (Chapter 15) f = 7. Then.96 Hz Correct Part E Find the period of the traveling waves. ANSWER: v max = 2. close the video window and answer the question on the right. ANSWER: v = 14. ANSWER: T = 0.7 m/s Correct Part G Find the maximum transverse speed of a point on the string.80 m/s Correct Video Tutor: OutofPhase Speakers First.126 s Correct Part F Find the speed of the traveling waves. You will be asked to use your knowledge of physics to predict the outcome of an experiment. You can watch the video again at any point. Will the sound be loudest at the nodes or antinodes of the standing wave pattern? Where will there be no sound? ANSWER: 0λ λ 3 4 1 2 1 4 λ λ λ Correct Problem 15. and they each emit a sound of wavelength λ . how far from the leftmost speaker should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces.370 N . Hint 1. One speaker is 180∘ out of phase with respect to the other. The center of gravity of this beam is onethird of the way along the beam from the end where wire A is attached. each 1.5λ .50 A 1740N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B). Note that the speakers represent antinodes of the resulting standing wave pattern. Ignore the effect of the weight of the wires on the tension in the wires. If we separate the speakers by a distance 1. what is the time delay between the arrival of the two pulses at the ceiling? Express your answer with the appropriate units. 32/33 .Homework 4 (Chapter 15) Part A Two speakers face each other. Part A If you pluck both strings at the same time at the beam. How to approach the problem. Select all that apply. Make a sketch of the situation.21 m long and weighing 0. 0 Hz .0 N is suspended from the lower end of a thin wire that is 3. since that wire has the greater tension. When the rock is totally submerged in a liquid. the fundamental frequency for the wire is 25. ANSWER: ρ = 1890 kg m3 Correct 33/33 .71 A large rock that weighs 164.Homework 4 (Chapter 15) ANSWER: tB − tA = 2. with the top of the rock just below the surface. When the rock is in air. The mass of the wire is small enough that its effect on the tension in the wire can be neglected. The density of the rock is 3200 kg/m3 . Part A What is the density of the liquid? Express your answer with the appropriate units. so the pulse in wire A arrives first.0 Hz .60 ms Correct Part B Which pulse arrives first? ANSWER: in wire A in wire B Correct The wave pulse travels faster in wire A.00 m long. The upper end of the wire is held fixed. Problem 15. the fundamental frequency for transverse standing waves on the wire is 39.