10/13/2017 Mastering HW 1Mastering HW 1 Due: 11:59pm on Friday, September 29, 2017 To understand how points are awarded, read the Grading Policy for this assignment. Cosine Wave The graph shows the position x of an oscillating object as a function of time t. The equation of the graph is x(t) = A cos (ωt + ϕ), where A is the amplitude, ω is the angular frequency, and ϕ is a phase constant. The quantities M , N , and T are measurements to be used in your answers. Part A What is A in the equation? Hint 1. Maximum of x(t) What is the maximum value of x on the graph and what is the maximum of x(t) as described by the equation? The equation is just a constant multiplied by a cosine function. Cosine can only range from −1 to 1. ANSWER: T M 2M M /T T /2 Correct Part B What is ω in the equation? Hint 1. Period Think of the simpler equation x = cos (ωt) . The period T is the same as before. What does x equal when t = T ? Use the result to solve for ω. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588 1/29 10/13/2017 Mastering HW 1 T M 2πT 2π/T 2/T 1/T Correct Part C What is ϕ in the equation? Hint 1. Using the graph and trigonometry What is x equal to when t = −N ? Use your result for ω to solve for ϕ in terms of T , M , and N . Hint 2. Using the graph and Part B You might be able to find ϕ in terms of ω and then use your result from Part B. ANSWER: N T −N 2πN /T −2πN /T arccos(2πN /T ) Correct Good Vibes: Introduction to Oscillations Learning Goal: To learn the basic terminology and relationships among the main characteristics of simple harmonic motion. Motion that repeats itself over and over is called periodic motion. There are many examples of periodic motion: the earth revolving around the sun, an elastic ball bouncing up and down, or a block attached to a spring oscillating back and forth. The last example differs from the first two, in that it represents a special kind of periodic motion called simple harmonic motion. The conditions that lead to simple harmonic motion are as follows: There must be a position of stable equilibrium. There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object's displacement from its equilibrium position. Mathematically, the restoring force F ⃗ is given by F ⃗ = −kx,⃗ where x⃗ is the displacement from equilibrium and k is a constant that depends on the properties of the oscillating system. The resistive forces in the system must be reasonably small. In this problem, we will introduce some of the basic quantities that describe oscillations and the relationships among them. Consider a block of mass m attached to a spring with force constant k, as shown in the figure. The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at x = 0. If the block is pulled to the right a distance A and then released, A will be the amplitude of the resulting oscillations. Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588 2/29 doubled. the block has gained some speed.10/13/2017 Mastering HW 1 Part A After the block is released from x = A . temporarily coming to rest at x = −A. In SI units. it will ANSWER: remain at rest. it still pulls the block to the left. that is. It will. the frequency is ANSWER: unchanged. What is its frequency f ? Express your answer in hertz. denoted f . Correct Part C An oscillating object takes 0. halved. pass the equilibrium position and keep moving. Correct As the block begins its motion to the left. and the block will slow down. completing one cycle of motion. The spring will now be pushing the block to the right. if. ANSWER: https://session. is the number of cycles that are completed per unit of time: f = 1/T . the period is denoted T and is measured in seconds. The frequency. move to the left until it reaches x = −A and stop there. compressing the spring. After x = −A is reached. the block will begin its motion to the right. The time it takes the block to complete one cycle is called the period. its period is 0. Usually. the motion will repeat indefinitely. or hertz (Hz).com/myct/assignmentPrintView?assignmentID=5701588 3/29 . it accelerates. Although the restoring force decreases as the block approaches equilibrium. so by the time the equilibrium position is reached. pushed by the spring.10 s. as we've assumed. move to the left until it reaches equilibrium and stop there. The block will pass the equilibrium position and continue until it reaches x = A. The motion will then repeat. there is no friction. Part B If the period is doubled. therefore. f is measured in inverse seconds.masteringphysics. move to the left until it reaches x = −A and then begin to move to the right.10 s to complete one cycle. what is the period T ? Express your answer in seconds.masteringphysics. Note that the vertical axis represents the x coordinate of the oscillating object.com/myct/assignmentPrintView?assignmentID=5701588 4/29 .10/13/2017 Mastering HW 1 f = 10 Hz Correct Part D If the frequency is 40 Hz. Part E Which points on the x axis are located a distance A from the equilibrium position? ANSWER: R only Q only both R and Q Correct Part F Suppose that the period is T . and the horizontal axis represents time. Which of the following points on the t axis are separated by the time interval T ? ANSWER: K and L K and M K and P L and N M and P https://session.025 s Correct The following questions refer to the figure that graphically depicts the oscillations of the block on the spring. ANSWER: T = 0. masteringphysics. Part G What is the period T ? Express your answer in seconds.10/13/2017 Mastering HW 1 Correct Now assume for the remaining Parts G . Hint 1. ANSWER: d = 0. that the x coordinate of point R is 0.36 m Correct Harmonic Oscillator Kinematics https://session.005 s .0050 s.48 m Correct Part J What distance d does the object cover between the moments labeled K and N on the graph? Express your answer in meters.12 m and the t coordinate of point K is 0. Then you can set aT = 0. Dividing by the fraction a will give the period T .J. what fraction of a full wavelength is covered? Call that fraction a. ANSWER: d = 0. ANSWER: t = 0. How to approach the problem In moving from the point t = 0 to the point K.01 s Correct Part I What distance d does the object cover during one period of oscillation? Express your answer in meters. ANSWER: T = 0.com/myct/assignmentPrintView?assignmentID=5701588 5/29 .02 s Correct Part H How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? Express your answer in seconds. The block is slowly pulled from its equilibrium position to some position xinit > 0 along the x axis. The length of the relaxed spring is L. the initial position of the block. and ω (Greek letter omega). S. What about S? To find the relationship between S and other variables. The block rests on a frictionless horizontal surface. Hint 2. Part A Using the general equation for x(t) given in the problem introduction. The constant C in this case is simply xinit . Consider t = 0 Evaluate the general expression for x(t) when t = 0 . the velocity of the block is zero. ω. the block is released with zero initial velocity. The goal is to determine the position of the block x(t) as a function of time in terms of ω and xinit . therefore.com/myct/assignmentPrintView?assignmentID=5701588 6/29 . obtain the expression for the block's velocity v(t) in terms of C .10/13/2017 Mastering HW 1 Learning Goal: To understand the application of the general harmonic equation to the kinematics of a spring oscillator. The equilibrium position of the left side of the block is defined to be x = 0. where C . and t. let us consider another initial condition that we know: At t = 0. Hint 2. Then evaluate the general expression for v(t) when t = 0. How to approach the problem Using the general equation x(t). ANSWER: xinit = C Correct This result is a good first step. Your task. S.masteringphysics. Hint 1. It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t) = C cos (ωt) + S sin (ωt) . and ω are constants. Some useful trigonometry Recall that cos 0 = 1 and sin 0 = 0 . Differentiating harmonic functions Recall that https://session. Part B Find the value of S using the given condition that the initial velocity of the block is zero: v(0) = 0 . One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. At time t = 0 . Hint 1. is to determine the values of C and S in terms of ω and xinit . express the initial position of the block xinit in terms of C . S. Hint 1. ANSWER: xinit tan(ωt) xinit ω xinit 0 Correct Part C What is the equation x(t) for the block? Express your answer in terms of t. https://session. ANSWER: x(t) = xinit cos(ωt) Correct In this problem the initial velocity is zero. The initial position of the block is the same as before. so the quantity |xinit | is the maximum displacement of the block from the equilibrium position.com/myct/assignmentPrintView?assignmentID=5701588 7/29 . the block's starting position is given by xnew (t = 0) = L + xinit . the formula for x(t) can be rewritten as x(t) = A cos ωt . Part D Find the equation for the block's position xnew (t) in the new coordinate system.masteringphysics.10/13/2017 Mastering HW 1 d cos ωt = −ω sin ωt dt and d sin ωt = ω cos ωt . ω. often denoted A. but in the new coordinate system. ω (Greek letter omega). Using this notation. The magnitude of the maximum displacement is called the amplitude. Now. dt Note the negative sign in the first formula. and t. Express your answer in terms of L. imagine that we have exactly the same physical situation but that the x axis is translated. and xinit . so that the position of the wall is now defined to be x = 0 . xinit . Start with the general solution Use the general solution and the values for C and S obtained in the previous parts. Therefore. Equilibrium position Changing the origin of the coordinate system has no effect on the physical parameters of the problem (e. Part A Rank the time required for the crates to return to their initial positions from largest to smallest. Determining the mass At equilibrium. a ranking can be determined.g.com/myct/assignmentPrintView?assignmentID=5701588 8/29 . Hint 2.10/13/2017 Mastering HW 1 Hint 1. If m cannot be determined. The crates are released and the springs compress to a length L before bringing the crates back up to their original positions. the compression of the spring at equilibrium is one-half the total distance the crate falls before beginning to move back upward. The period of a mass-spring system is given by −− m T = 2π√ k . the frequency or the amplitude of the block's oscillations). that is. overlap them.. What is the difference. Since the crate oscillates with equal amplitude above and below the equilibrium position.masteringphysics. Hint 1. between xnew (t) in the new coordinate system and x(t) in the old coordinate system? ANSWER: xnew (t) − x(t) = L ANSWER: xnew (t) = L + xinit cos(ωt) Correct Period of a Mass-Spring System Ranking Task Different mass crates are placed on top of springs of uncompressed length L0 and stiffness k. Formula for the period The period is defined as the time it takes for an oscillator to go through one complete cycle of its motion. the time for each crate to return to its initial position is one period. the force of the spring upward is equal to the force of gravity downward: ksequilibrium = mg. at any moment. Solving for the mass we get ksequilibrium m = g . The only difference is that now the block is oscillating around x = L whereas before it was oscillating around x = 0. the ranking cannot be determined based on the information provided. https://session. To rank items as equivalent. if m can be determined from the provided information. Rank from largest to smallest. The initial velocity is still zero. Therefore. Hint 3. and substituting mass into the period formula. if the answer "B to D" were correct. https://session. Expressing smaxiμm in terms of known quantities. For example. will allow you to determine the correct ranking. and Acceleration of an Oscillator Learning Goal: To learn to find kinematic variables from a graph of position vs.10/13/2017 Mastering HW 1 1 sequilibrium = 2 smaximum . Some of the questions ask you to determine ranges on the graph over which a statement is true. ANSWER: Reset Help largest smallest The correct ranking cannot be determined. choose the most complete answer. The graph of the position of an oscillating object as a function of time is shown. Velocity. then "B to C" would technically also be correct--but you will only recieve credit for choosing the most complete answer. L0 is the uncompressed length of the spring and L is the maximum compression of the spring. Correct Position. When answering these questions. The total distance the crate falls before beginning to move back upward is given by smaximum = L0 − L . Determining smaxiμm As defined in the problem.com/myct/assignmentPrintView?assignmentID=5701588 9/29 . Combining these two ideas results in ksmaximum m = 2g . time.masteringphysics. com/myct/assignmentPrintView?assignmentID=5701588 10/29 .masteringphysics.10/13/2017 Mastering HW 1 Part A Where on the graph is x > 0? ANSWER: A to B A to C C to D C to E B to D A to B and D to E Correct Part B Where on the graph is x < 0? ANSWER: A to B A to C C to D C to E B to D A to B and D to E Correct Part C Where on the graph is x = 0? ANSWER: A only C only E only A and C A and C and E B and D Correct Part D Where on the graph is the velocity v > 0 ? https://session. dt ANSWER: https://session. t curve is zero: = 0.masteringphysics. dx(t) v(t) = .com/myct/assignmentPrintView?assignmentID=5701588 11/29 . How to tell if v = 0 dx(t) The velocity is zero when the slope of the x vs. dt Thus. When is the slope greater than 0 on this graph? ANSWER: A to B A to C C to D C to E B to D A to B and D to E Correct Part E Where on the graph is the velocity v < 0 ? ANSWER: A to B A to C C to D C to E B to D A to B and D to E Correct Part F Where on the graph is the velocity v = 0 ? Hint 1.10/13/2017 Mastering HW 1 Hint 1. Finding instantaneous velocity Instantaneous velocity is the derivative of the position function with respect to time. t graph. you can find the velocity at any time by calculating the slope of the x vs. t graph. Where is the curvature greater than 0? ANSWER: A to B A to C C to D C to E B to D A to B and D to E Correct Part H Where on the graph is the acceleration a < 0 ? ANSWER: A to B A to C C to D C to E B to D A to B and D to E https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588 12/29 . Finding acceleration Acceleration is the second derivative of the position function with respect to time: 2 d x(t) a= 2 .10/13/2017 Mastering HW 1 A only B only C only D only E only A and C A and C and E B and D Correct Part G Where on the graph is the acceleration a > 0 ? Hint 1. The acceleration of a curve is negative for downward curvature and positive for upward curvature. dt This means that the sign of the acceleration is the same as the sign of the curvature of the x vs. Inflection points are where the curvature of the graph changes sign. How to tell if a = 0 The acceleration is zero at the inflection points of the x vs.10/13/2017 Mastering HW 1 Correct Part I Where on the graph is the acceleration a = 0 ? Hint 1. ANSWER: A only B only C only D only E only A and C A and C and E B and D Correct Problem 14. t graph. ANSWER: A = 10 cm Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588 13/29 .5 Part A What is the amplitude of the oscillation shown in the figure? Express your answer to two significant figures and include the appropriate units. In other words.masteringphysics. In this problem.10/13/2017 Mastering HW 1 Part B What is the frequency of the oscillation shown in the figure? Express your answer to two significant figures and include the appropriate units.50 Hz Correct Part C What is the phase constant of the oscillation shown in the figure? Express your answer to two significant figures and include the appropriate units. Answer the following questions. 1 K= mv 2 . it can be excluded from the calculations. Note that. For such a system. Assume that the force constant k. A.com/myct/assignmentPrintView?assignmentID=5701588 14/29 . the potential energy is stored in the spring and is given by 1 U = kx 2 . a common example of a harmonic oscillator is a mass attached to a spring. https://session. We will also assume that there are no resistive forces. or harmonic oscillators. ANSWER: ϕ 0 = 120 ∘ Correct Energy of Harmonic Oscillators Learning Goal: To learn to apply the law of conservation of energy to the analysis of harmonic oscillators. Such an analysis can be simplified if one assumes that mechanical energy is not dissipated. C. labeled A. E = constant . K is the kinetic energy. one can analyze many aspects of motion of the oscillator. are given. and D. ANSWER: f = 0. E = K + U = constant. 2 where k is the force constant of the spring and x is the distance from the equilibrium position. since the gravitational potential energy is not changing in this case. the mass of the block. that is. 2 where m is the mass of the block and v is the speed of the block. obey the law of conservation of energy just like all other systems do. and the amplitude of vibrations. where E is the total mechanical energy of the system. and U is the potential energy. The kinetic energy of the system is. Systems in simple harmonic motion. as shown in the figure . we will consider a horizontally moving block attached to a spring. Using energy considerations. B. as always. Consider a harmonic oscillator at four different moments. As you know. m. When is the speed at a minimum? Keep in mind that speed is the magnitude of the 1 mv 2 velocity. How does the velocity change? 2 Recall that K = . where x is the distance from equilibrium.com/myct/assignmentPrintView?assignmentID=5701588 15/29 .masteringphysics. Thus. the greater the potential energy. the farther the block is from equilibrium. where v is the speed of the block. so the lowest value that it can take is zero. ANSWER: A B C D https://session. Consider the position of the block Recall that U = 1 2 kx 2 .10/13/2017 Mastering HW 1 Part A Which moment corresponds to the maximum potential energy of the system? Hint 1. When is the block farthest from equilibrium? ANSWER: A B C D Correct Part B Which moment corresponds to the minimum kinetic energy of the system? Hint 1. Recall that E . ANSWER: https://session. At that moment. K . moving to the right. 1 = kA = Kmin = 0 = K +U 2 1 2 E= kA .masteringphysics. and the block is momentarily at rest. moving toward equilibrium.com/myct/assignmentPrintView?assignmentID=5701588 16/29 . Part C Consider the block in the process of oscillating. the acceleration of the block—decrease. As the block approaches equilibrium. Consider the distance from equilibrium The smallest potential energy corresponds to the smallest distance from equilibrium. Therefore. the mechanical energy of a harmonic oscillator equals its potential energy at the maximum or minimum displacement. it gains speed. at the amplitude displacement. 2 Therefore. ANSWER: at the equilibrium position. therefore. of course. the block must be moving to the left. the force applied by the spring —and. moving away from equilibrium. 2 In general.10/13/2017 Mastering HW 1 Correct When the block is displaced a distance A from equilibrium. the maximum potential energy is U max . the spring is stretched (or compressed) the most. Correct Part D Which moment corresponds to the maximum kinetic energy of the system? Hint 1. At what position does the object begin to slow down? ANSWER: A B C D Correct Part E Which moment corresponds to the minimum potential energy of the system? Hint 1. If the kinetic energy of the block is increasing. Consider the velocity of the block As the block begins to move away from the amplitude position. The speed of the block is at a maximum when the acceleration becomes zero. 1 = 2 ANSWER: A B C D Correct Part G Find the kinetic energy K of the block at the moment labeled B. m Part F At which moment is K = U ? Hint 1. Recall that 1 2 E = K +U and that U = 0 at the equilibrium position. 1 2 E= kA . 2 we can now conclude that 1 2 1 2 kA = mvmax . Consider the potential energy At this moment. Meanwhile. Hint 1. U U max . 2 Recalling what we found out before.10/13/2017 Mastering HW 1 A B C D Correct When the block is at the equilibrium position. How to approach the problem Find the potential energy first. Hint 2. Find the potential energy Find the potential energy U of the block at the moment labeled B. the block is at its maximum speed (vmax ). then use conservation of energy. of course. Express your answer in terms of k and A.com/myct/assignmentPrintView?assignmentID=5701588 17/29 . 1 2 E= mvmax . The maximum kinetic energy can then be written as Kmax = mv2 max . U = Umin = 0 . the spring is not stretched (or compressed) at all. Therefore.masteringphysics. 2 2 or −− k vmax = √ A = ωA. At that moment. https://session. Use the formula for Umax to obtain the corresponding distance from equilibrium. since the initial coordinate is zero. S. The length of the relaxed spring is L. the block is released with zero initial velocity. Physical laws Combine the expressions for Hooke's law F = −kx and Newton's 2nd law F = ma .com/myct/assignmentPrintView?assignmentID=5701588 18/29 .and xinit and then use the connection between x(t) and a(t) to find the acceleration. m. is to determine the values of C . At time t = 0 . and ω are constants. Express your answer in terms of k. Part A Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. ANSWER: k a(t) = − m x(t) https://session. Hint 1. the deformation of the spring at any time equals the coordinate of the block x(t). S. The block rests on a frictionless horizontal surface. It is known that a general solution for the position of a harmonic oscillator is x(t) = C cos (ωt) + S sin (ωt) . The other end is attached to a block of mass m. therefore. The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k. One end of a spring with spring constant k is attached to the wall. and xinit . and the coordinate of the block x(t). and ω in terms of k. m. Note that. The equilibrium position of the left side of the block is defined to be x = 0. m.10/13/2017 Mastering HW 1 Express your answer in terms of kand A. where C . Your task.masteringphysics. The block is slowly pulled from its equilibrium position to some position xinit > 0 along the x axis. ANSWER: 2 U = 1 kA 8 ANSWER: 2 K = 3 kA 8 Correct Harmonic Oscillator Acceleration Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time. the period of oscillations T depend only on the intrinsic physical characteristics of the system ( k and m). As a result. the block is displaced to the right and when it is released.com/myct/assignmentPrintView?assignmentID=5701588 19/29 . where it is again pulled back toward equilibrium. while the left end of the spring is held fixed. Express your answer in terms of ω. t. a force acts on it to pull it back toward equilibrium. m Compare these expressions to determine ω. When the spring is neither compressed nor stretched. ± Introduction to Simple Harmonic Motion Consider the system shown in the figure. undergoing oscillations.10/13/2017 Mastering HW 1 Correct The negative sign in the answer is important: It indicates that the restoring force (the tension of the spring) is always directed opposite to the block's displacement. the block moves back and forth from one side of the equilibrium position to the other. and x(t). Frequency and period do not depend on the initial conditions or the amplitude of the motion. to the left--and vice versa. you obtained two expressions for a(t): k k a(t) = − x(t) = − (C cos(ωt) + S sin(ωt)) m m and k 2 a(t) = − x(t) = −ω (C cos(ωt) + S sin(ωt)). If the spring is stretched. Using the previous results In the previous parts. so it overshoots. An important example of periodic motion is simple harmonic motion (SHM) and we will use the mass-spring system described here to introduce some of its properties. Hint 1.masteringphysics. that is. Since we are ignoring friction (a good approximation to many cases). Part B Using the fact that acceleration is the second derivative of position. Express your answer in terms of k and m. It consists of a block of mass m attached to a spring of negligible mass and force constant k. Part A https://session. stopping somewhere on the other side. The resulting oscillations take the name of periodic motion. the mechanical energy of the system is conserved and the oscillations repeat themselves over and over. ANSWER: −− ω = √ k m Correct Note that the angular frequency ω and. The motion that we have just described is typical of most systems when they are displaced from equilibrium and experience a restoring force that tends to bring them back to their equilibrium position. therefore. the restoring force is pulling back. The block is free to move on a frictionless horizontal surface. ANSWER: a(t) = 2 −ω x(t) Correct Part C Find the angular frequency ω. the block is in equilibrium. By the time the block has returned to the equilibrium position. When the block is pulled to the right from the equilibrium position. it has picked up some kinetic energy. find the acceleration of the block a(t) as a function of time. ANSWER: The restoring force is constant. The negative sign expresses the fact that the force exerted by the spring acts in the direction opposite the direction in which the displacement has occurred. (The positive direction is to the right. a0 . Also note that the spring exerts a varying force that is proportional to displacement. Express your answer in terms of some or all of the variables A. and k. Note that if the block is displaced a certain distance from its equilibrium position. Hint 1. Forces exerted on the block in the x direction https://session.masteringphysics. Part B As shown in the figure. Hooke's law The expression known as Hooke's law says that a spring stretched or compressed by a distance x exerts a force given by F = −kx.com/myct/assignmentPrintView?assignmentID=5701588 20/29 .10/13/2017 Mastering HW 1 Which of the following statements best describes the characteristic of the restoring force in the spring-mass system described in the introduction? Hint 1. the resulting periodic motion is referred to as simple harmonic motion. m. Hint 1. a coordinate system with the origin at the equilibrium position is chosen so that the x coordinate represents the displacement from the equilibrium position. when the block is released at a distance A from its equilibrium position? Express your answer in terms of some or all of the variables A. when the block is at a distance A from its equilibrium position. Correct Whenever the oscillations are caused by a restoring force that is directly proportional to displacement. Find the restoring force Find Fx . The restoring force is directly proportional to the displacement of the block. The restoring force is maximum when the block is in the equilibrium position. m. where k is a constant characteristic of the spring called the spring constant. Find which force is the restoring force Which of the following forces plays the role of the restoring force? ANSWER: gravity friction the force exerted by the spring the normal force Hint 2.) What is the initial acceleration of the block. and k. The restoring force is proportional to the mass of the block. the x component of the net force acting on the block. the spring is stretched by the same distance. that is. What do you expect the block's acceleration will be when the block is to the left of its equilibrium position and has undergone a negative displacement? Part D Select the correct expression that gives the block's acceleration at a distance x from the equilibrium position. the block's acceleration is not constant. an object in equilibrium does not accelerate. How to approach the problem Hooke's law gives you an expression for the force F exerted on the mass at a given displacement. you can find a formula for the acceleration of the mass attached to the spring. and k. A characteristic of equilibrium By definition. Using this equation. since all the other forces (gravity and the normal force) act in the vertical direction. where a is the acceleration and m is the mass. ANSWER: Fx = −kA ANSWER: k a0 = − m A Correct Part C What is the acceleration a1 of the block when it passes through its equilibrium position? Express your answer in terms of some or all of the variables A. its acceleration is always opposite in sign with respect to displacement. the block can be either to the right or left of its equilibrium position. Moreover. instead. Hint 1.10/13/2017 Mastering HW 1 The x component of the net force acting on the block is due exclusively to the force exerted by the spring. the acceleration's magnitude decreases to zero as the block goes through its equilibrium position.com/myct/assignmentPrintView?assignmentID=5701588 21/29 . ANSWER: a = −kx a = kx k a= x m k a=− x m Correct Whether the block undergoes a positive or negative displacement. Hint 1. Then.masteringphysics. Note that x can be either positive or negative. m. This is a fundamental property of simple harmonic motion. it is directly proportional to displacement. Newton's 2nd law tells you that a = F /m. ANSWER: a1 = 0 Correct Your results from Parts B and C show that the acceleration of the block is negative when the block has undergone a positive displacement. https://session. The speed of the block is zero when it is between its rightmost position and the equilibrium position. the location where the block's acceleration changes sign must also be the location where its speed reaches its maximum value. between its rightmost position and the equilibrium position. that is. Part E Hint 1. between its leftmost position and the equilibrium position. at either its rightmost or leftmost position. its acceleration decreases from positive values to negative values. its speed increases from zero to a certain value and then decreases back to zero. How to approach the problem When the block is in motion. Since the acceleration is directly proportional to displacement. In particular. select the correct phrases to complete the following statements. between its leftmost position and the equilibrium position. ANSWER: The magnitude of the block's acceleration reaches its maximum value when the block is in the equilibrium position.masteringphysics. where it stops increasing and starts to decrease. you found that a = −(k/m)x . This means that as the block moves away from its rightmost position toward its leftmost position. ANSWER: https://session. How to approach the problem As the block moves from its rightmost position to its leftmost position. when the direction of motion changes. ANSWER: in the equilibrium position. How to approach the problem In Part D. its speed can be zero only when its velocity changes sign.10/13/2017 Mastering HW 1 Using the information found so far. Correct Part F Hint 1. at either its rightmost or leftmost position. Correct Part G Hint 1. it must reach its maximum value when displacement is maximum.com/myct/assignmentPrintView?assignmentID=5701588 22/29 . How to find the equation for acceleration To determine the correct equation for the acceleration. Vertical Mass-and-Spring Oscillator A block of mass m is attached to the end of an ideal spring. v x = −B sin ωt ax = −C cos ωt Correct Further calculations would show that the constants B and C can be expressed in terms of A and ω. where ω is a constant characteristic of the system. Correct Part H Because of the periodic properties of SHM. Mathematically. vx and ax . v x = −B cos ωt ax = −C cos ωt . you will have a set of values of t at which x = A.10/13/2017 Mastering HW 1 in the equilibrium position. v x = −B sin ωt ax = C cos ωt . recall that when x = A the speed of the block is zero. rather. you will not find a unique value for t . if the block is released at a distance A from its equilibrium position. For example. its displacement x varies with time t according to the equation x = A cos ωt. The spring has an unknown spring constant k. Hint 1. at either its rightmost or leftmost position. ANSWER: . ω must be expressed in radians per second so that the quantity ωt is expressed in radians. between the leftmost position and the equilibrium position. Due to the weight of the block.masteringphysics. At this point you simply need to determine which function among ±B sin ωt and ±B cos ωt is zero at those calculated values of t. v x = B cos ωt ax = C sin ωt . In the expressions below. You can verify then whether your result is correct by calculating the acceleration at t = 0 and comparing it with your result in Part B. respectively. Use this equation and the information you now have on the acceleration and speed of the block as it moves back and forth from one side of its equilibrium position to the other to determine the correct set of equations for the block's x components of velocity and acceleration. How to find the equation for velocity To determine the correct equation for the velocity. you can calculate when x = A from the given equation for displacement. https://session. If time is measured is seconds. B and C are nonzero positive constants. Hint 2. the block remains at rest when the spring is stretched a distance h from its equilibrium length. The speed of the block reaches its maximum value when the block is between the rightmost position and the equilibrium position.com/myct/assignmentPrintView?assignmentID=5701588 23/29 . simply substitute the equation x = A cos ωt into the expression for a found in Part D and group all positive constants together. the mathematical equations that describe this motion involve sine and cosine functions. When you do that. and g . the force that the spring exerts on the block? ANSWER: Fs = hk Hint 2. the magnitude of the acceleration due to gravity.masteringphysics. Sum of forces acting on the block Since the block is not accelerating. Force due to spring What is Fs . Find the resulting angular frequency ω of the block's oscillation about its equilibrium position. h. the magnitude of the acceleration due to gravity. the gravitational force on the block? ANSWER: Fg = −mg ANSWER: ∑ Fy = 0 = hk − mg ANSWER: g k = m( ) h Correct Part B Suppose that the block gets bumped and undergoes a small vertical displacement.10/13/2017 Mastering HW 1 Part A What is the spring constant k? Express the spring constant in terms of given quantities and g . write an expression for the net vertical force ∑ Fy acting on the block. Express the sum of the vertical forces in terms of m. Hint 1. k. https://session.com/myct/assignmentPrintView?assignmentID=5701588 24/29 . Force due to gravity What is Fg . Hint 1. Taking the positive y direction to be upward. the net force acting on the block must be zero. Hence F = −w = −kx. However. ANSWER: k = 1200 N/m Correct Hint 3. and knowing the mass. By measuring h and ω (both fairly simple measurements). the weight of the object will be counteracted by the restoring force in the spring. these parameters are what would determine the extension h of the spring when the block is hanging: h = mg/k. https://session. which can be seen by drawing a force diagram of the fish on the spring. Hint 1. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.com/myct/assignmentPrintView?assignmentID=5701588 25/29 .masteringphysics. which might make little sense. Hint 1. where ω is the angular frequency and f = 2.00 Hz is the frequency.0 cm from the 0 to 180 N reading. Express your answer numerically in radians per second. ANSWER: − − g ω = √ h Correct It may seem that this result for the frequency does not depend on either the mass of the block or the spring constant. Formula for angular frequency −−−− The angular frequency of simple harmonic motion for a body of mass m acted on by a restoring force with force constant k is given by ω = √k/m . The scale has a length of 15.00 Hz . Hint 1. the spring constant can be calculated from this equation. then use this with the angular frequency of the bouncing fish to calculate its mass. Hint 2. for a given oscillation. Using the reading At rest. One way of thinking about this problem is to consider both k and g as unknowns. Calculate the spring constant Calculate the spring constant k for the spring in the fish scale. How to approach the problem Calculate the spring constant for the fish scale. what is the mass m of the fish? Express your answer in kilograms. ω = 2πf . you can determine the value of the spring constant and the acceleration due to gravity experimentally. Relating frequency and angular frequency Recall that. ± The Fish Scale A vertical scale on a spring balance reads from 0 to 180 N . Because we know both the maximum weight the scale can show and the length the spring is stretched at that weight.10/13/2017 Mastering HW 1 Express the frequency in terms of given quantities and g . Express your answer in newtons per meter. the magnitude of the acceleration due to gravity. Calculate the angular frequency Calculate the angular frequency ω for the fish oscillating on the spring. Part A Ignoring the mass of the spring. Hint 1. 6 radians/s Correct Hint 4. ANSWER: m = 7. approximately what will the pendulum's new period be? Hint 1. Part A If the bob's mass is doubled. Since the gravitational acceleration appears in the denominator.masteringphysics. Formula for the angular frequency of a mass on a spring −−−− An object of mass m on the end of a spring with spring constant k will oscillate with frequency ω = √k/m .60 kg Correct Changing the Period of a Pendulum A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T .10/13/2017 Mastering HW 1 ANSWER: ω = 12. ANSWER: T /2 T √2T 2T Correct Part B If the pendulum is brought on the moon where the gravitational acceleration is about g/6. where g is the acceleration due to gravity. Period of a simple pendulum The period T of a simple pendulum of length L is given by −− L T = 2π√ g . the period must increase when the gravitational acceleration decreases. ANSWER: https://session. How to approach the problem Recall the formula of the period of a simple pendulum. approximately what will its period now be? Hint 1.com/myct/assignmentPrintView?assignmentID=5701588 26/29 . It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall. Period of a Pendulum Ranking Task Part A Six pendulums of mass m and length L as shown are released from rest at the same angle θ from vertical. −− L T = 2π√ g . The roles of mass and length The force of gravity is the only force with a component along the direction of motion of the pendulum bob. Pendulums with larger lengths require that the bob attached to the pendulum must travel a larger distance before completing a cycle. giving them the same acceleration. These objects are said to be in free fall. Hint 1.10/13/2017 Mastering HW 1 T /6 T /√6 √6T 6T Correct Part C If the pendulum is taken into the orbiting space station what will happen to the bob? Hint 1. It will oscillate much faster with a period that approaches zero. Because this force provides the same acceleration to objects regardless of mass. where T is the period (the time for one complete cycle). all of the masses will accelerate at the same rate when released. The frequency of an oscillator is the number of complete cycles it goes through in a given time interval. Correct In the space station. the tension in the string is zero and the bob does not fall relative to the point to which the string is attached. where all objects undergo the same acceleration due to the earth's gravity. The gravitational force acts to bring the bob back to its equilibrium position. To rank items as equivalent. overlap them. ANSWER: It will continue to oscillate in a vertical plane with the same period. the earth's gravity acts on both the station and everything inside it. The frequency f is the inverse of the period.masteringphysics. Hint 2. Frequency For a simple pendulum. Rank the pendulums according to the number of complete cycles of motion each pendulum goes through per minute. L is the length of the pendulum. In the space station. It will no longer oscillate because there is no gravity in space. How to approach the problem Recall that the oscillations of a simple pendulum occur when a pendulum bob is raised above its equilibrium position and let go. ANSWER: https://session. Rank from most to least complete cycles of motion per minute. causing the pendulum bob to fall. and g is the acceleration due to gravity.com/myct/assignmentPrintView?assignmentID=5701588 27/29 . com/myct/assignmentPrintView?assignmentID=5701588 28/29 . Hint 1.10/13/2017 Mastering HW 1 Reset Help most least The correct ranking cannot be determined. ANSWER: https://session.0 full swing cycles in a time of 133 s . ANSWER: T = 1.41 s Incorrect. Try Again. a space explorer constructs a simple pendulum of length 52. Correct ± Gravity on Another Planet After landing on an unfamiliar planet. 4 attempts remaining Hint 3. The explorer finds that the pendulum completes 94. and use this to calculate the magnitude of the gravitational acceleration on the planet. Express your answer in seconds. Calculate the period Calculate the period T of the pendulum. Hint 2.0 cm .masteringphysics. Equation for the period −−−−−−− The period of a simple pendulum is given by the equation T = 2π√L/g planet . Part A What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second. where L is the length of the pendulum and g planet is the magnitude of the gravitational acceleration on the planet. How to approach the problem Calculate the period of the pendulum. https://session.10/13/2017 Mastering HW 1 g planet = 10. You received 48.6%.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588 29/29 .56 out of a possible total of 53 points.3 m/s 2 Correct Score Summary: Your score on this assignment is 91.