Chapter 10: Simple Harmonic Motion

Transcription

1 Chapter 10: Simple Harmonic Motion Oscillations are back-and forth motions. Sometimes the word vibration is used in place of oscillation; for our purposes, we can consider the two words to represent the same class of motions. In the ideal case of no friction, free oscillations are a sub-class of periodic motions; that is, in the absence of friction, all free oscillations are periodic motions, but not all periodic motions are oscillations. For example, uniform circular motion (which we studied earlier in this course) is periodic, but not considered an oscillation. Uniform circular motion can be modelled by sine or cosine functions of time (think back to the unit circle in high-school math when you were learning trigonometry). (Sine and cosine functions are collectively known as sinusoidal functions, or sinusoids for short.) If the restoring force that causes oscillation is a linear function of position, then the resulting oscillatory motion can also be modelled by a sinusoidal function of time. There is a close relationship between uniform circular motion and oscillatory motion caused by a linear restoring force; we won't explore this, but check the textbook (Section 10.2) if you're interested. What is a restoring force? What is a linear restoring force? Ch10 Page 1

2 Notice that the formula for Hooke's law is represented by the graph; the magnitude of the restoring force is proportional to the magnitude of the displacement from equilibrium, and in the opposite direction. The constant of proportionality is the stiffness constant of the spring. Q: What are the units of k? What are some typical values for the stiffness constant for coil springs in your experience (ones in your car's shock absorbers, in your Ch10 Page 2

3 ball-point pen, attached to your aluminum door, etc.)? Here is an example position-time diagram for an oscillation: Notice that the position-time diagram for the oscillation resembles the graph of a sinusoidal function. You'll get a chance to see that this must be so for an oscillator that is subject to a linear restoring force (Hooke's law) later in the chapter. Now is a good time to review sinusoidal functions, so let's do it: Ch10 Page 3

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5 Exercises Ch10 Page 5

6 1. Determine the amplitude, period, frequency, and angular frequency for each function. In each case, time t is measured in seconds and displacement x is measured in centimetres. (a) (b) c. (d) 2. Sketch a graph of each position function in parts (a), (b), and (c) of Exercise Calculus lovers only! Determine a formula for the derivative of the sine function, and a formula for the derivative of the cosine function, valid for angles measured in degrees. Ch10 Page 6

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9 Exercises 4. Consider an oscillation with position function x = 20 cos (4t) where x is measured in cm and t is measured in s. (a) Determine the positions on the x-axis that are turning points. (b) Determine the times at which the oscillator is at the turning points. (c) Determine the times at which the oscillator is at the Ch10 Page 9

10 equilibrium position. (d) Determine the times at which the speed of the oscillator is at (i) a maximum, and (ii) a minimum. (e) Determine the times at which the acceleration of the oscillator is at (i) a maximum, and (ii) a minimum. Ch10 Page 10

11 Note that the period (and therefore also both the frequency and angular frequency) does not depend on the amplitude of the oscillation. This is interesting. Does this match with your experience? Exercises 5. What would the position-time graphs look like for an oscillator that is released from several different starting amplitudes? 6. Consider an oscillator of mass 4 kg attached to a spring with stiffness constant 200 N/m. The mass is pulled to an initial amplitude of 5 cm and then released. (a) Determine the angular frequency, frequency, and period of the oscillation. (b) Write a position-time function for the oscillator. 7. A block of mass 3.2 kg is attached to a spring. The resulting position-time function of this oscillator is x = 23.7 sin(4.3t), where t is measured in seconds. Determine the stiffness of the spring. Elastic potential energy An argument similar in spirit to the one above for gravitational potential energy shows that the formula for elastic potential energy is (see page 285ff in Section 10.3 of the textbook for details; calculus lovers, antidifferentiate Hooke's law) where x is the displacement of the spring from its equilibrium position and k is the stiffness constant of the spring. Where is potential energy stored? Answering this question leads Ch10 Page 11

12 to the concept of a force field; we'll discuss force fields in PHYS 1P22/1P92. Ch10 Page 12

13 Exercises Ch10 Page 13

14 8. Consider a block of mass 5.1 kg attached to a spring. The position-time function of this oscillator is x = 8.2 sin(2.7t), where x is measured in cm and t is measured in seconds. (a) Determine the total mechanical energy of the oscillator. (b) Determine the stiffness constant of the spring. (c) Determine the maximum potential energy. (d) Determine the maximum kinetic energy. (e) Determine the positions at which the kinetic energy and the potential energy of the oscillator are equal. Pendulum motion Recall from earlier in these lecture notes that for a block on the end of a spring, applying Newton's law to the block results in This is an example of a differential equation, and to solve a differential equation means to determine a position function x(t) that satisfies the equation. You'll learn how to do this in second-year physics (PHYS 2P20), and second-year math (MATH 2P08), but for now you can verify that position functions of the form Ch10 Page 14

15 and both satisfy the differential equation above, provided that satisfies a certain condition (the same condition that was observed earlier in the notes). From a different perspective, we can also infer that if a physical phenomenon is modelled by a differential equation of the form given above, the phenomenon is an example of simple harmonic motion. Let's consider a simple pendulum. First draw a free-body diagram: In the radial direction, applying Newton's second law gives: In the tangential direction, applying Newton's second law gives: Ch10 Page 15

16 This looks very similar to the differential equation written earlier that represents simple harmonic motion. But not exactly; if the sine theta were replaced by just theta, then the equation would have the form of the SHM differential equation. Note that for small angles, the value of sine theta is about equal to theta, provided that theta is measured in radians: Ch10 Page 16

17 Construct a table of values using a calculator and you'll see for yourself that sine theta is nearly equal to theta for small angles, provided that theta is measured in radians. The approximation becomes increasingly better the smaller the angle is. Also note that as long as theta is measured in radians; this is an example of a power series, which you'll learn about later in MATH 1P06 or MATH 1P02, if you are taking either of these two courses. For small values of theta, higher order terms are very small, and so this formula embodies the approximation given above. Thus, for a pendulum with a small amplitude (so that the angle is always small), the motion is approximately SHM, described by the differential equation Substituting the model Ch10 Page 17

18 into the differential equation leads to expressions for the period and frequency: Substituting this expression into the differential equation gives: Thus Ch10 Page 18

19 and therefore Ch10 Page 19

20 Swinging your arms while running or walking; note how the period of the swing is modified by changing the effective length of your arms: Ch10 Page 20

21 Damped oscillations Cars have shock absorbers to make the ride smoother. A shock absorber consists of a stiff spring together with a damping tube. (The damping tube consists of a piston in an enclosed cylinder that is filled with a thick (i.e., viscous) oil.) Without the damping tube, a car would oscillate for a long time after going over a bump in a road; the damping tube helps to limit both the amplitude and duration of the oscillations. When the damping tube doesn't work anymore, the car tends to oscillate for a long time after going over a bump, which is annoying. The same thing happens with a screen door when its damping tube malfunctions. Ch10 Page 21

22 (There is a spring attached to the door, but it is not shown in the photograph.) The position function for a damped oscillator is modified as follows: You can think of this position function as representing a sort of sinusoid, but one with a variable amplitude; the amplitude is the constant A times the exponential factor, which steadily decreases as time passes. Ch10 Page 22

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24 The damping tube in a screen door is adjustable. If the resistance is too great, then the door will take a long time to shut after it is opened. If the resistance is too little, then the door will swing back and forth many times before shutting. There is an ideal medium amount of resistance ("critical damping") which works best; you'll learn more about this, and see how these three cases (overdamping, critical damping, and underdamping) follow naturally from different classes of solutions to the appropriate differential equation, in PHYS 2P20 and MATH 2P08. Ch10 Page 24

25 Resonance Examples of driven oscillations: child sloshing back and forth in a bathtub adult pushing a child on a playground swing Some systems have a natural oscillating frequency; if you drive the system at its natural oscillating frequency, its amplitude can increase dramatically. This phenomenon is called resonance. In many situations, one tries to avoid resonant oscillations. For example, that annoying vibration in your dashboard when you are driving on the highway at a certain speed is very annoying. More seriously, soldiers are trained to break ranks when they march across a bridge, because if their collective march is at the same frequency as the natural frequency of the bridge, then there is danger that they could collapse the bridge. (This is an ancient custom, from an age when many bridges were made of wood.) Ch10 Page 25

26 Engineers must be careful to design bridges and tall buildings so that they don't have natural vibration frequencies; otherwise an unlucky wind could cause dangerous largeamplitude vibrations. Ch10 Page 26

27 Another good example of resonance is tuned electrical circuits, such as the ones used in radio or television reception. Radio waves of many different frequencies are incident on a radio receiver in your home; each tries to "drive" electrical oscillations in an electrical circuit. The natural frequency of the electrical circuit can be adjusted so that it will resonate with only a certain frequency of radio wave; this is how you "tune in" to a certain radio station. The oscillations due to the resonant frequency persist, while all the other frequencies are rapidly damped. 1. Answers to exercises from earlier in the notes Determine the amplitude, period, frequency, and angular frequency for each function. In each case, time t is measured in seconds and displacement x is measured in centimetres. Ch10 Page 27

28 frequency for each function. In each case, time t is measured in seconds and displacement x is measured in centimetres. (a) (b) (c) (d) Solution: (a) A = 4 cm, T = s, f = 1/ cycles/s, = 2 rad/s (b) A = 3 cm, T = s, f = 1/2 cycles/s, = rad/s (c) A = 1/2 cm, T = 4 s, f = 1/(4 cycles/s, = 1/2 rad/s (d) This is not a sinusoidal function, and not a periodic function either. Therefore this function does not have an amplitude, a period, a frequency, or an angular frequency; it cannot serve as a model for a simple harmonic motion. 2. Sketch a graph of each position function in parts (a), (b), and (c) of Exercise 1. Solution: Ch10 Page 28

29 3. Calculus lovers only! Determine a formula for the derivative of the sine function, and a formula for the derivative of the cosine function, valid for angles measured in degrees. [See me for a discussion on this point it's a bit complicated.] 4. Consider an oscillation with position function x = 20 cos (4t) where x is measured in cm and t is measured in s. (a) Determine the positions on the x-axis that are turning points. (b) Determine the times at which the oscillator is at the turning points. (c) Determine the times at which the oscillator is at the Ch10 Page 29

30 equilibrium position. (d) Determine the times at which the speed of the oscillator is at (i) a maximum, and (ii) a minimum. (e) Determine the times at which the acceleration of the oscillator is at (i) a maximum, and (ii) a minimum. Solution: (a) x = 20 cm (b) t = 0 s, /4 s, /2 s, 3 /4 s, s, etc. (c) t = 0 s, /8 s, 3 /8 s, 5 /8 s, 7 /8 s, etc. (d) The speed of the oscillator is maximum at the times in Part (c) (i.e., when the oscillator passes through the equilibrium position); the speed of the oscillator is minimum (i.e., zero) at the times in Part (b) (i.e., when the oscillator is at a turning point). (e) The accelerator of the oscillator is maximum at the times in Part (b) (i.e., when the oscillator is at a turning point); the accelerator of the oscillator is minimum (i.e., zero) at the times in Part (c) (i.e., when the oscillator passes through the equilibrium position). 5. What would the position-time graphs look like for an oscillator that is released from several different starting amplitudes? Solution: The stiffness constant of the spring and the value of the mass are the same, so the frequency and period of each motion are the same. The only difference is the amplitude. Ch10 Page 30

31 6. Consider an oscillator of mass 4 kg attached to a spring with stiffness constant 200 N/m. The mass is pulled to an initial amplitude of 5 cm and then released. (a) Determine the angular frequency, frequency, and period of the oscillation. (b) Write a position-time function for the oscillator. Solution: 7. A block of mass 3.2 kg is attached to a spring. The resulting position-time function of this oscillator is x = 23.7 sin(4.3t), where t is measured in seconds. Determine the stiffness of the spring. Solution: 8. Consider a block of mass 5.1 kg attached to a spring. The position-time function of this oscillator is x = 8.2 sin(2.7t), where x is measured in cm and t is measured in seconds. Ch10 Page 31

32 where x is measured in cm and t is measured in seconds. (a) Determine the total mechanical energy of the oscillator. (b) Determine the stiffness constant of the spring. (c) Determine the maximum potential energy. (d) Determine the maximum kinetic energy. (e) Determine the positions at which the kinetic energy and the potential energy of the oscillator are equal. Solution: Ah, we don't know the value of the stiffness constant k, so we should calculate it: Thus, (e) When the kinetic and potential energies are equal, each is half the total energy. Therefore, the potential energy is equal to half of the total energy at the positions of interest. Thus, Ch10 Page 32

33 Additional Solved Problems Example: An object in simple harmonic motion has an amplitude of 6.0 cm and a frequency of 0.50 Hz. Draw a position-time graph showing two cycles of the motion. Solution: Should we use a sine model or a cosine model? It doesn't really matter, so I'll use a sine model. Because the amplitude is 6.0 cm, the positiontime graph looks like this: We just have to correctly label the horizontal axis now. We are given that the frequency is 0.5 Hz, so we can determine the period: Thus, one complete cycle of the position-time graph takes 2.0 s. Ch10 Page 33

34 We weren't asked for a formula, but using the general formula we can write a specific formula for the position function of this motion: Example: Some passengers on an ocean cruise may suffer from motion sickness as the ship rocks back and forth on the waves. At one position on the ship, passengers experience a vertical motion of amplitude 1 m with a period of 15 s. (a) To one significant figure, what is the maximum acceleration of the passengers during this motion? (b) What fraction is this of g? Solution: This time I'll use a cosine model; the reasoning would be the same for a sine model. Ch10 Page 34

35 Because the cosine function takes values between -1 and 1, the maximum value of the acceleration is related to the right side of the equation on the previous line by We are given that the amplitude A is 1 m. If we can determine the angular frequency, then we will be able to calculate the maximum value of the acceleration. But we are told the period, so we will be able to determine the angular frequency: Thus, the maximum value of the acceleration is (b) In terms of the acceleration due to gravity, the maximum acceleration of the ship's oscillation is The maximum acceleration is only about 2% of the acceleration due to gravity, so presumably it's not too bad on the ship. Ch10 Page 35

36 Example: (a) When the displacement of a mass on a spring is (1/2)A, what fraction of the mechanical energy is kinetic energy and what fraction is potential energy? (b) At what displacement, as a fraction of A, is the mechanical energy half kinetic and half potential? Solution: First determine a formula for the total energy of the oscillator. In general, However, this doesn't seem to help much, because we also need a way of comparing how much of the total energy is of each type. What to do? Well, note that when x = A, the oscillator momentarily stops, so its kinetic energy is zero. Thus, the total energy at this position is But remember, the total energy of the oscillator is conserved (there is no friction). Thus, the total energy of the oscillator has the same value throughout its entire motion. Thus, the ratio of the potential energy to the total energy is Thus, when the displacement is half of the amplitude, Ch10 Page 36

37 Thus, when the displacement is half of the amplitude, Thus, when the displacement is half of the amplitude, the potential energy is 1/4 of the total energy, and therefore the kinetic energy is 3/4 of the total energy. (b) Using the solution from Part (a), Thus, the energy is shared equally between kinetic energy and potential energy when the displacement is plus-or-minus about 70% of the amplitude. Ch10 Page 37

38 Example: A 1.0 kg block is attached to a spring with stiffness constant 16 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40 cm/s. Determine (a) the amplitude of the subsequent oscillations, and (b) the block's speed at the position where x = (1/2)A. Solution: From the given information we can immediately calculate the angular frequency. I am guessing this might be useful, so let's do that first: Because the block is at its equilibrium position and is given an initial speed, a sine model is appropriate for its position function: We are told that the oscillator's initial speed, which is also its maximum speed, is 40 cm/s. Thus, (b) As we learned in a previous problem, the total energy can be expressed in terms of the amplitude as follows: Ch10 Page 38

39 (b) As we learned in a previous problem, the total energy can be expressed in terms of the amplitude as follows: To determine the block's speed when the amplitude is half of its maximum, Example: A 507 g mass oscillates with an amplitude of 10.0 cm on a spring Ch10 Page 39

40 whose stiffness constant is 20.0 N/m. Determine (a) the period, (b) the maximum speed, and (c) the total energy. Solution: (b) The maximum speed occurs when x = 0: Ch10 Page 40

41 Example: On your first trip to planet X you happen to take along a 200 g mass, a 40.0-cm-long spring, a metre stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on Earth, but you can't find this information in your visitor's guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You measure that the mass stretches the spring by 31.2 cm. You then pull the mass down an additional 10.0 cm and release it. With your stopwatch you measure that 10 oscillations take 14.5 s. Determine the value of the acceleration due to gravity on Planet X. Solution: Let g represent the acceleration due to gravity on Planet X. Placing the mass on the spring stretches it by 31.2 cm, which is the same as m. This gives us a relation between g and the stiffness constant k of the spring, which you can get by drawing a free body diagram and using Newton's second law: Now that we have a relation between g and k, we see that we can calculate g provided that we can determine k independently of this equation. Re-reading the statement of the problem, looking for unused information, it occurs to us that perhaps we can use the fact that the oscillator completes 10 oscillations in 14.5 s to determine k. Indeed, this can be done: Ch10 Page 41

42 Therefore, Example: A 1.00 kg block is attached to a horizontal spring with stiffness constant 2500 N/m. The block is at rest on a frictionless surface. A 10.0-g bullet is fired into the block in the face opposite the spring, and it sticks. (a) Determine the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm. (b) Could you determine the bullet's speed by measuring the oscillation frequency? If so, how? If not, why not? Strategy: When the bullet strikes the block, mechanical energy is not conserved. That is, some of the initial kinetic energy of the bullet is converted to sound, thermal energy, deformation of the bullet and block, and so on. Ch10 Page 42

43 However, momentum is conserved in this process. Once the bullet has embedded into the block, mechanical energy is conserved in the subsequent motion of the block + bullet system, because there is no friction between the block and the surface on which it rests. Thus: Use conservation of momentum to determine the speed of the bullet + block just after the bullet is embedded. Then use conservation of mechanical energy to relate the speed just determined to the amplitude of the subsequent oscillations. Solution: (a) The following diagrams are intended to make this strategy clear: Momentum is conserved in the collision process; that is, the momentum just before and just after the collision is conserved: Ch10 Page 43

44 After the bullet + block begin to oscillate, mechanical energy is conserved. Thus, the initial kinetic energy of the system is equal to the maximum potential energy stored in the spring: Thus, the speed of the bullet is (b) No. The frequency of the oscillation depends only on the stiffness constant of the spring and the total mass of the bullet + block. The oscillation frequency is independent of the bullet's speed, so measuring the oscillation frequency can't possibly give us any insight into the bullet's speed. The greater the bullet's speed, the greater the amplitude of the oscillation; thus, it's the amplitude of the oscillation that is connected to the speed of the bullet. Ch10 Page 44

45 Example: A spring is hung from the ceiling. A kg block is then attached to the free end of the spring. When released from rest, the block drops m before momentarily coming to rest, after which it moves back upward. (a) Determine the spring's stiffness constant. (b) Determine the block's angular frequency of oscillation. Solution: See page 279 of the textbook, which explains clearly how to think about the unstretched length of the spring and the amplitude of the resulting oscillations. Carefully examine the following diagram: When the block reaches the lowest point of its motion, the loss in the block's gravitational potential energy is balanced by the gain in the spring's elastic potential energy: The block's angular frequency is therefore Ch10 Page 45

46 Example: A vertical spring with stiffness constant 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30-kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, determine the height (in cm) above the compressed spring from which the block was dropped. Solution: By the principle of conservation of energy, the gravitational potential energy of the block "before" is converted to elastic potential energy in Ch10 Page 46

47 the spring "after." Thus, Example: A simple pendulum is swinging back and forth through a small angle, its motion repeating every 1.25 s. How much longer should the pendulum be made so that its period increases by 0.20 s? Solution: Ch10 Page 47

48 Thus, the pendulum should be made 35% longer to increase its period by 0.2 s. Ch10 Page 48