Chapter 15 Serway - 8th Edition PDF

Oscillatory Motion chapter 15 15.1 Motion of an Object Attached to a Spring 15.2 Analysis Model: Particle in Simple Harmonic Motio...

Oscillatory Motion chapter 15 15.1 Motion of an Object Attached to a Spring 15.2 Analysis Model: Particle in Simple Harmonic Motion 15.3 Energy of the Simple Harmonic Oscillator 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 15.5 The Pendulum 15.6 Damped Oscillations 15.7 Forced Oscillations Periodic motion is motion of an object that regularly returns to a given position after a fixed time interval. With a little thought, we can identify several types of periodic motion in everyday life. Your car returns to the driveway each afternoon. You return to the dinner table each night to eat. A bumped chandelier swings back and forth, returning to the same posi- tion at a regular rate. The Earth returns to the To reduce swaying in tall buildings because of the wind, tuned dampers are placed near the top of the building. These mechanisms include an object of same position in its orbit around the Sun each large mass that oscillates under computer control at the same frequency as year, resulting in the variation among the four the building, reducing the swaying. The 730-ton suspended sphere in the photograph above is part of the tuned damper system of the Taipei Financial seasons. Center, at one time the world’s tallest building. (Ranjit Doroszkeiwicz/Alamy) In addition to these everyday examples, numerous other systems exhibit periodic motion. The molecules in a solid oscillate about their equilibrium positions; electromag- netic waves, such as light waves, radar, and radio waves, are characterized by oscillating electric and magnetic field vectors; and in alternating-current electrical circuits, voltage, current, and electric charge vary periodically with time. A special kind of periodic motion occurs in mechanical systems when the force acting on an object is proportional to the position of the object relative to some equilibrium position. If this force is always directed toward the equilibrium position, the motion is called simple harmonic motion, which is the primary focus of this chapter. 434  15.1 | Motion of an Object Attached to a Spring 435 15.1 Motion of an Object Attached to a Spring As a model for simple harmonic motion, consider a block of mass m attached to the end of a spring, with the block free to move on a frictionless, horizontal surface (Active Fig. 15.1). When the spring is neither stretched nor compressed, the block is at rest at the position called the equilibrium position of the system, which we identify as x 5 0 (Active Fig. 15.1b). We know from experience that such a system oscillates back and forth if disturbed from its equilibrium position. We can understand the oscillating motion of the block in Active Figure 15.1 qualitatively by first recalling that when the block is displaced to a position x, the spring exerts on the block a force that is proportional to the position and given by Hooke’s law (see Section 7.4): Fs 5 2kx (15.1) W Hooke’s law We call Fs a restoring force because it is always directed toward the equilibrium position and therefore opposite the displacement of the block from equilibrium. That is, when the block is displaced to the right of x 5 0 in Active Figure 15.1a, the position is positive and the restoring force is directed to the left. When the block is displaced to the left of x 5 0 as in Figure 15.1c, the position is negative and the restoring force is directed to the right. When the block is displaced from the equilibrium point and released, it is a particle under a net force and consequently undergoes an acceleration. Applying Newton’s second law to the motion of the block, with Equation 15.1 providing the net force in the x direction, we obtain 2kx 5 max k ax 5 2 x (15.2) m Pitfall Prevention 15.1 That is, the acceleration of the block is proportional to its position, and the direc- The Orientation of the Spring tion of the acceleration is opposite the direction of the displacement of the block Active Figure 15.1 shows a horizontal spring, with an attached block slid- from equilibrium. Systems that behave in this way are said to exhibit simple har- ing on a frictionless surface. Another monic motion. An object moves with simple harmonic motion whenever its accel- possibility is a block hanging from eration is proportional to its position and is oppositely directed to the displacement a vertical spring. All the results we from equilibrium. discuss for the horizontal spring If the block in Active Figure 15.1 is displaced to a position x 5 A and released are the same for the vertical spring with one exception: when the block from rest, its initial acceleration is 2kA/m. When the block passes through the equi- is placed on the vertical spring, its librium position x 5 0, its acceleration is zero. At this instant, its speed is a maxi- weight causes the spring to extend. mum because the acceleration changes sign. The block then continues to travel to If the resting position of the block is the left of equilibrium with a positive acceleration and finally reaches x 5 2A, at defined as x 5 0, the results of this which time its acceleration is 1kA/m and its speed is again zero as discussed in Sec- chapter also apply to this vertical system. tions 7.4 and 7.9. The block completes a full cycle of its motion by returning to the S Fs When the block is displaced to the right of equilibrium, a m the force exerted by the x spring acts to the left. x x0 S Fs  0 When the block is at its equilibrium position, the b m force exerted by the spring x is zero. S Fs x0 When the block is displaced c m to the left of equilibrium, ACTIVE FIGURE 15.1 x the force exerted by the x spring acts to the right. A block attached to a spring moving x0 on a frictionless surface. 436 CHAPTER 15 | Oscillatory Motion original position, again passing through x 5 0 with maximum speed. Therefore, the block oscillates between the turning points x 5 6A. In the absence of friction, this idealized motion will continue forever because the force exerted by the spring is conservative. Real systems are generally subject to friction, so they do not oscillate forever. We shall explore the details of the situation with friction in Section 15.6. Quick Quiz 15.1 A block on the end of a spring is pulled to position x 5 A and released from rest. In one full cycle of its motion, through what total dis- tance does it travel? (a) A/2 (b) A (c) 2A (d) 4A 15.2 Analysis Model: Particle in Simple Harmonic Motion The motion described in the preceding section occurs so often that we identify the particle in simple harmonic motion model to represent such situations. To develop a mathematical representation for this model, we will generally choose x as the axis along which the oscillation occurs; hence, we will drop the subscript-x notation in this discussion. Recall that, by definition, a 5 dv/dt 5 d 2x/dt 2, so we can express Equation 15.2 as d 2x k 2 5 2 m x (15.3) dt If we denote the ratio k/m with the symbol v2 (we choose v2 rather than v so as to Pitfall Prevention 15.2 make the solution we develop below simpler in form), then A Nonconstant Acceleration The acceleration of a particle in k simple harmonic motion is not con- v2 5 (15.4) m stant. Equation 15.3 shows that its acceleration varies with position x. and Equation 15.3 can be written in the form Therefore, we cannot apply the kine- matic equations of Chapter 2 in this d 2x situation. 5 2v 2x (15.5) dt 2 Let’s now find a mathematical solution to Equation 15.5, that is, a function x(t) that satisfies this second-order differential equation and is a mathematical repre- sentation of the position of the particle as a function of time. We seek a function whose second derivative is the same as the original function with a negative sign and multiplied by v2. The trigonometric functions sine and cosine exhibit this behavior, so we can build a solution around one or both of them. The following cosine function is a solution to the differential equation: Position versus time for X x 1 t 2 5 A cos 1 vt 1 f 2 (15.6) a particle in simple harmonic motion where A, v, and f are constants. To show explicitly that this solution satisfies Equa- tion 15.5, notice that dx d 5 A cos 1 vt 1 f 2 5 2vA sin 1 vt 1 f 2 (15.7) dt dt d 2x d Pitfall Prevention 15.3 5 2vA sin 1 vt 1 f 2 5 2v 2A cos 1 vt 1 f 2 (15.8) Where’s the Triangle? dt 2 dt Equation 15.6 includes a trigonomet- Comparing Equations 15.6 and 15.8, we see that d 2x/dt 2 5 2v2x and Equation 15.5 ric function, a mathematical function that can be used whether it refers is satisfied. to a triangle or not. In this case, the The parameters A, v, and f are constants of the motion. To give physical signifi- cosine function happens to have the cance to these constants, it is convenient to form a graphical representation of the correct behavior for representing motion by plotting x as a function of t as in Active Figure 15.2a. First, A, called the the position of a particle in simple amplitude of the motion, is simply the maximum value of the position of the parti- harmonic motion. cle in either the positive or negative x direction. The constant v is called the angu-  15.2 | Analysis Model: Particle in Simple Harmonic Motion 437 lar frequency, and it has units1 of radians per second. It is a measure of how rapidly x the oscillations are occurring; the more oscillations per unit time, the higher the T A value of v. From Equation 15.4, the angular frequency is t k v5 (15.9) –A Åm a The constant angle f is called the phase constant (or initial phase angle) and, x along with the amplitude A, is determined uniquely by the position and velocity A of the particle at t 5 0. If the particle is at its maximum position x 5 A at t 5 0, the phase constant is f 5 0 and the graphical representation of the motion is as t shown in Active Figure 15.2b. The quantity (vt 1 f) is called the phase of the –A motion. Notice that the function x(t) is periodic and its value is the same each time vt increases by 2p radians. b Equations 15.1, 15.5, and 15.6 form the basis of the mathematical representation of the particle in simple harmonic motion model. If you are analyzing a situation ACTIVE FIGURE 15.2 and find that the force on an object modeled as a particle is of the mathematical (a) An x–t graph for a particle under- form of Equation 15.1, you know the motion is that of a simple harmonic oscillator going simple harmonic motion. The amplitude of the motion is A, and and the position of the particle is described by Equation 15.6. If you analyze a sys- the period (defined in Eq. 15.10) is tem and find that it is described by a differential equation of the form of Equation T. (b) The x–t graph for the special 15.5, the motion is that of a simple harmonic oscillator. If you analyze a situation case in which x 5 A at t 5 0 and and find that the position of a particle is described by Equation 15.6, you know the hence f 5 0. particle undergoes simple harmonic motion. x Quick Quiz 15.2 Consider a graphical representation (Fig. 15.3) of simple harmonic motion as described mathematically in Equation 15.6. When the t particle is at point 훽 on the graph, what can you say about its position and 훽 velocity? (a) The position and velocity are both positive. (b) The position and velocity are both negative. (c) The position is positive, and the velocity Figure 15.3 (Quick Quiz 15.2) An is zero. (d) The position is negative, and the velocity is zero. (e) The position x–t graph for a particle undergoing is positive, and the velocity is negative. (f) The position is negative, and the simple harmonic motion. At a par- velocity is positive. ticular time, the particle’s position is indicated by 훽 in the graph. Quick Quiz 15.3 Figure 15.4 shows two curves representing particles under- x going simple harmonic motion. The correct description of these two motions t is that the simple harmonic motion of particle B is (a) of larger angular fre- quency and larger amplitude than that of particle A, (b) of larger angular Particle A frequency and smaller amplitude than that of particle A, (c) of smaller angu- x lar frequency and larger amplitude than that of particle A, or (d) of smaller angular frequency and smaller amplitude than that of particle A. t Let us investigate further the mathematical description of simple harmonic Particle B motion. The period T of the motion is the time interval required for the particle to go through one full cycle of its motion (Active Fig. 15.2a). That is, the values of x Figure 15.4 (Quick Quiz 15.3) Two and v for the particle at time t equal the values of x and v at time t 1 T. Because the x–t graphs for particles undergoing simple harmonic motion. The ampli- phase increases by 2p radians in a time interval of T, tudes and frequencies are different for the two particles. [v(t 1 T) 1 f] 2 (vt 1 f) 5 2p 1We have seen many examples in earlier chapters in which we evaluate a trigonometric function of an angle. The argument of a trigonometric function, such as sine or cosine, must be a pure number. The radian is a pure number because it is a ratio of lengths. Angles in degrees are pure numbers because the degree is an artificial “unit”; it is not related to measurements of lengths. The argument of the trigonometric function in Equation 15.6 must be a pure number. Therefore, v must be expressed in radians per second (and not, for example, in revolutions per second) if t is expressed in seconds. Furthermore, other types of functions such as logarithms and exponential functions require arguments that are pure numbers. 438 CHAPTER 15 | Oscillatory Motion Pitfall Prevention 15.4 Simplifying this expression gives vT 5 2p, or Two Kinds of Frequency 2p We identify two kinds of frequency T5 (15.10) v for a simple harmonic oscillator: f, called simply the frequency, is mea- The inverse of the period is called the frequency f of the motion. Whereas the sured in hertz, and v, the angular period is the time interval per oscillation, the frequency represents the number of frequency, is measured in radians per oscillations the particle undergoes per unit time interval: second. Be sure you are clear about which frequency is being discussed 1 v or requested in a given problem. f5 5 (15.11) T 2p Equations 15.11 and 15.12 show the relationship between the two The units of f are cycles per second, or hertz (Hz). Rearranging Equation 15.11 gives frequencies. 2p v 5 2pf 5 (15.12) T Equations 15.9 through 15.11 can be used to express the period and frequency of the motion for the particle in simple harmonic motion in terms of the character- istics m and k of the system as 2p m Period X T5 5 2p (15.13) v Åk 1 1 k Frequency X f5 5 (15.14) T 2pÅ m That is, the period and frequency depend only on the mass of the particle and the force constant of the spring and not on the parameters of the motion, such as A or f. As we might expect, the frequency is larger for a stiffer spring (larger value of k) and decreases with increasing mass of the particle. We can obtain the velocity and acceleration2 of a particle undergoing simple harmonic motion from Equations 15.7 and 15.8: dx Velocity of a particle in simple X v5 5 2vA sin 1 vt 1 f 2 (15.15) harmonic motion dt d 2x Acceleration of a particle in X a5 5 2v 2A cos 1 vt 1 f 2 (15.16) simple harmonic motion dt 2 From Equation 15.15, we see that because the sine and cosine functions oscillate between 61, the extreme values of the velocity v are 6vA. Likewise, Equation 15.16 shows that the extreme values of the acceleration a are 6v2A. Therefore, the maxi- mum values of the magnitudes of the velocity and acceleration are k Maximum magnitudes of X v max 5 vA 5 A (15.17) velocity and acceleration in Åm simple harmonic motion k a max 5 v 2A 5 A (15.18) m Figure 15.5a plots position versus time for an arbitrary value of the phase con- stant. The associated velocity–time and acceleration–time curves are illustrated in Figures 15.5b and 15.5c, respectively. They show that the phase of the velocity dif- fers from the phase of the position by p/2 rad, or 908. That is, when x is a maxi- mum or a minimum, the velocity is zero. Likewise, when x is zero, the speed is a maximum. Furthermore, notice that the phase of the acceleration differs from the phase of the position by p radians, or 1808. For example, when x is a maximum, a has a maximum magnitude in the opposite direction. 2Because the motion of a simple harmonic oscillator takes place in one dimension, we denote velocity as v and accel- eration as a, with the direction indicated by a positive or negative sign as in Chapter 2.  15.2 | Analysis Model: Particle in Simple Harmonic Motion 439 x T Quick Quiz 15.4 An object of mass m is hung from a spring and set into oscil- lation. The period of the oscillation is measured and recorded as T. The xi A object of mass m is removed and replaced with an object of mass 2m. When t this object is set into oscillation, what is the period of the motion? a (a) 2T (b) !2 T (c) T (d) T/ !2 (e) T/2 v Equation 15.6 describes simple harmonic motion of a particle in general. Let’s vi vmax now see how to evaluate the constants of the motion. The angular frequency v is t evaluated using Equation 15.9. The constants A and f are evaluated from the ini- tial conditions, that is, the state of the oscillator at t 5 0. b Suppose a block is set into motion by pulling it from equilibrium by a distance A a and releasing it from rest at t 5 0 as in Active Figure 15.6. We must then require our a max solutions for x(t) and v(t) (Eqs. 15.6 and 15.15) to obey the initial conditions that t x(0) 5 A and v(0) 5 0: x(0) 5 A cos f 5 A c v(0) 5 2vA sin f 5 0 Figure 15.5 Graphical representa- tion of simple harmonic motion. These conditions are met if f 5 0, giving x 5 A cos vt as our solution. To check this (a) Position versus time. (b) Velocity solution, notice that it satisfies the condition that x(0) 5 A because cos 0 5 1. versus time. (c) Acceleration versus time. Notice that at any specified The position, velocity, and acceleration of the block versus time are plotted in time the velocity is 908 out of phase Figure 15.7a for this special case. The acceleration reaches extreme values of 7v2A with the position and the accelera- when the position has extreme values of 6A. Furthermore, the velocity has extreme tion is 1808 out of phase with the values of 6vA, which both occur at x 5 0. Hence, the quantitative solution agrees position. with our qualitative description of this system. Let’s consider another possibility. Suppose the system is oscillating and we define t 5 0 as the instant the block passes through the unstretched position of the spring x0 A while moving to the right (Active Fig. 15.8). In this case, our solutions for x(t) and t0 xi  A v(t) must obey the initial conditions that x(0) 5 0 and v(0) 5 vi: m vi  0 x(0) 5 A cos f 5 0 v(0) 5 2vA sin f 5 vi ACTIVE FIGURE 15.6 The first of these conditions tells us that f 5 6p/2. With these choices for f, the A block–spring system that begins its second condition tells us that A 5 7vi/v. Because the initial velocity is positive and motion from rest with the block at the amplitude must be positive, we must have f 5 2p/2. Hence, the solution is x 5 A at t 5 0. vi p x5 cos avt 2 b v 2 x T 3T x T 3T 2 2 2 T 2 t t T t0 v v T 3T xi  0 x0 2 2 v  vi t t S vi T T 3T T 2 m 2 a a T 3T 2 T 2 t t T T 3T 2 2 ACTIVE FIGURE 15.8 a b The block–spring system is undergo- ing oscillation, and t 5 0 is defined Figure 15.7 (a) Position, velocity, and acceleration versus time for the block in Active Figure 15.6 at an instant when the block passes under the initial conditions that at t 5 0, x(0) 5 A, and v(0) 5 0. (b) Position, velocity, and accelera- through the equilibrium position tion versus time for the block in Active Figure 15.8 under the initial conditions that at t 5 0, x(0) 5 0, x 5 0 and is moving to the right with and v(0) 5 vi. speed vi. 440 CHAPTER 15 | Oscillatory Motion The graphs of position, velocity, and acceleration versus time for this choice of t 5 0 are shown in Figure 15.7b. Notice that these curves are the same as those in Figure 15.7a, but shifted to the right by one-fourth of a cycle. This shift is described mathe- matically by the phase constant f 5 2p/2, which is one-fourth of a full cycle of 2p. Ex a m pl e 15.1 A Block–Spring System A 200-g block connected to a light spring for which the force constant is 5.00 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced 5.00 cm from equilibrium and released from rest as in Active Figure 15.6. (A) Find the period of its motion. SOLUTION Conceptualize Study Active Figure 15.6 and imagine the block moving back and forth in simple harmonic motion once it is released. Set up an experimental model in the vertical direction by hanging a heavy object such as a stapler from a strong rubber band. Categorize The block is modeled as a particle in simple harmonic motion. We find values from equations developed in this section for the particle in simple harmonic motion model, so we categorize this example as a substitution problem. k 5.00 N/m Use Equation 15.9 to find the angular frequency of the v5 5 5 5.00 rad/s Å m Å 200 3 1023 kg block–spring system: 2p 2p Use Equation 15.13 to find the period of the system: T5 5 5 1.26 s v 5.00 rad/s (B) Determine the maximum speed of the block. SOLUTION Use Equation 15.17 to find v max: v max 5 vA 5 (5.00 rad/s)(5.00 3 1022 m) 5 0.250 m/s (C) What is the maximum acceleration of the block? SOLUTION Use Equation 15.18 to find a max: a max 5 v2A 5 (5.00 rad/s)2(5.00 3 1022 m) 5 1.25 m/s2 (D) Express the position, velocity, and acceleration as functions of time in SI units. SOLUTION Find the phase constant from the initial condition that x(0) 5 A cos f 5 A S f 5 0 x 5 A at t 5 0: Use Equation 15.6 to write an expression for x(t): x 5 A cos (vt 1 f) 5 0.050 0 cos 5.00t Use Equation 15.15 to write an expression for v(t): v 5 2vA sin (vt 1 f) 5 20.250 sin 5.00t Use Equation 15.16 to write an expression for a(t): a 5 2v2A cos (vt 1 f) 5 21.25 cos 5.00t WHAT IF? What if the block were released from the same initial position, xi 5 5.00 cm, but with an initial velocity of vi 5 20.100 m/s? Which parts of the solution change, and what are the new answers for those that do change? Answers Part (A) does not change because the period is independent of how the oscillator is set into motion. Parts (B), (C), and (D) will change. Write position and velocity expressions for the initial (1) x(0) 5 A cos f 5 xi conditions: (2) v(0) 5 2vA sin f 5 vi  15.2 | Analysis Model: Particle in Simple Harmonic Motion 441 15.1 cont. 2vA sin f vi Divide Equation (2) by Equation (1) to find the phase 5 A cos f x i constant: vi 20.100 m/s tan f 5 2 52 5 0.400 vx i 1 5.00 rad/s 2 1 0.050 0 m 2 f 5 tan21 (0.400) 5 0.121p xi 0.050 0 m Use Equation (1) to find A: A5 5 5 0.053 9 m cos f cos 1 0.121p 2 Find the new maximum speed: v max 5 vA 5 (5.00 rad/s)(5.39 3 1022 m) 5 0.269 m/s Find the new magnitude of the maximum acceleration: a max 5 v2A 5 (5.00 rad/s)2(5.39 3 1022 m) 5 1.35 m/s2 Find new expressions for position, velocity, and accelera- x 5 0.053 9 cos (5.00t 1 0.121p) tion in SI units: v 5 20.269 sin (5.00t 1 0.121p) a 5 21.35 cos (5.00t 1 0.121p) As we saw in Chapters 7 and 8, many problems are easier to solve using an energy approach rather than one based on variables of motion. This particular What If? is easier to solve from an energy approach. Therefore, we shall investigate the energy of the simple harmonic oscillator in the next section. Ex a m pl e 15.2 Watch Out for Potholes! A car with a mass of 1 300 kg is constructed so that its frame is supported by four springs. Each spring has a force con- stant of 20 000 N/m. Two people riding in the car have a combined mass of 160 kg. Find the frequency of vibration of the car after it is driven over a pothole in the road. SOLUTION Conceptualize Think about your experiences with automobiles. When you sit in a car, it moves downward a small dis- tance because your weight is compressing the springs further. If you push down on the front bumper and release it, the front of the car oscillates a few times. Categorize We imagine the car as being supported by a single spring and model the car as a particle in simple harmonic motion. Analyze First, let’s determine the effective spring constant of the four springs combined. For a given extension x of the springs, the combined force on the car is the sum of the forces from the individual springs. Find an expression for the total force on the car: F total 5 o (2kx) 5 2 ao k b x In this expression, x has been factored from the sum because it is the same for all four springs. The effective spring con- stant for the combined springs is the sum of the individual spring constants. Evaluate the effective spring constant: k eff 5 o k 5 4 3 20 000 N/m 5 80 000 N/m 1 keff 1 80 000 N/m Use Equation 15.14 to find the frequency of vibration: f5 5 5 1.18 Hz 2pÅ m 2pÅ 1 460 kg Finalize The mass we used here is that of the car plus the people because that is the total mass that is oscillating. Also notice that we have explored only up-and-down motion of the car. If an oscillation is established in which the car rocks back and forth such that the front end goes up when the back end goes down, the frequency will be different. continued 442 CHAPTER 15 | Oscillatory Motion 15.2 cont. WHAT IF? Suppose the car stops on the side of the road and the two people exit the car. One of them pushes down- ward on the car and releases it so that it oscillates vertically. Is the frequency of the oscillation the same as the value we just calculated? Answer The suspension system of the car is the same, but the mass that is oscillating is smaller: it no longer includes the mass of the two people. Therefore, the frequency should be higher. Let’s calculate the new frequency, taking the mass to be 1 300 kg: 1 keff 1 80 000 N/m f5 5 5 1.25 Hz 2pÅ m 2pÅ 1 300 kg As predicted, the new frequency is a bit higher. 15.3 Energy of the Simple Harmonic Oscillator Let us examine the mechanical energy of a system in which a particle undergoes simple harmonic motion, such as the block–spring system illustrated in Active Fig- ure 15.1. Because the surface is frictionless, the system is isolated and we expect the total mechanical energy of the system to be constant. We assume a massless spring, so the kinetic energy of the system corresponds only to that of the block. We can use Equation 15.15 to express the kinetic energy of the block as Kinetic energy of a simple X K 5 12mv 2 5 12mv 2A2 sin2 1 vt 1 f 2 (15.19) harmonic oscillator The elastic potential energy stored in the spring for any elongation x is given by 1 2 2 kx(see Eq. 7.22). Using Equation 15.6 gives Potential energy of a simple X U 5 12 kx 2 5 12 kA2 cos2 1 vt 1 f 2 (15.20) harmonic oscillator v2 We see that K and U are always positive quantities or zero. Because 5 k/m, we can express the total mechanical energy of the simple harmonic oscillator as E 5 K 1 U 5 12kA2 3 sin2 1 vt 1 f 2 1 cos2 1 vt 1 f 2 4 From the identity sin2 u 1 cos2 u 5 1, we see that the quantity in square brackets is unity. Therefore, this equation reduces to Total energy of a simple X E 5 12kA2 (15.21) harmonic oscillator That is, the total mechanical energy of a simple harmonic oscillator is a constant of the motion and is proportional to the square of the amplitude. The total mechani- cal energy is equal to the maximum potential energy stored in the spring when x 5 6A because v 5 0 at these points and there is no kinetic energy. At the equilibrium position, where U 5 0 because x 5 0, the total energy, all in the form of kinetic energy, is again 12kA2. Plots of the kinetic and potential energies versus time appear in Active Figure 15.9a, where we have taken f 5 0. At all times, the sum of the kinetic and potential energies is a constant equal to 12kA2, the total energy of the system. The variations of K and U with the position x of the block are plotted in Active Figure 15.9b. Energy is continuously being transformed between potential energy stored in the spring and kinetic energy of the block. Active Figure 15.10 illustrates the position, velocity, acceleration, kinetic energy, and potential energy of the block–spring system for one full period of the motion. Most of the ideas discussed so far are incorporated in this important figure. Study it carefully. Finally, we can obtain the velocity of the block at an arbitrary position by express- ing the total energy of the system at some arbitrary position x as  15.3 | Energy of the Simple Harmonic Oscillator 443 In either plot, notice that ACTIVE FIGURE 15.9 K  U  constant. (a) Kinetic energy and potential energy versus time for a simple 1 U K U  2 kx 2 K  12 mv 2 harmonic oscillator with f 5 0. (b) Kinetic energy and potential K, U K, U energy versus position for a simple harmonic oscillator. 1 2 1 2 2 kA 2 kA t x T T –A A 2 O a b E 5 K 1 U 5 12mv 2 1 12kx 2 5 12kA2 k 2 v56 1 A 2 x 2 2 5 6v "A2 2 x 2 (15.22) W Velocity as a function Åm of position for a simple When you check Equation 15.22 to see whether it agrees with known cases, you harmonic oscillator find that it verifies that the speed is a maximum at x 5 0 and is zero at the turning points x 5 6A. You may wonder why we are spending so much time studying simple harmonic oscillators. We do so because they are good models of a wide variety of physical phenomena. For example, recall the Lennard–Jones potential discussed in Exam- ple 7.9. This complicated function describes the forces holding atoms together. S amax % t x v a K U 100 a 50 1 2 0 0 A 0 v2A 0 2 kA S vmax % 100 b 50 T 0 vA 0 1 2 0 0 4 2 kA S amax % 100 c 50 T 1 2 A 0 v2A 0 2 kA 0 2 S vmax % 100 3T 1 2 d 50 0 vA 0 2 kA 0 0 4 S amax % 100 1 2 e 50 T A 0 v2A 0 2 kA 0 S v % 100 1 1 2 t x v v2x 2 mv 2 2 kx f 50 0 x Kinetic Potential Total energy energy energy x –A 0 –A ACTIVE FIGURE 15.10 (a) through (e) Several instants in the simple harmonic motion for a block–spring system. Energy bar graphs show the distribution of the energy of the system at each instant. The parameters in the table at the right refer to the block–spring system, assuming at t 5 0, x 5 A; hence, x 5 A cos vt. For these five special instants, one of the types of energy is zero. (f) An arbitrary point in the motion of the oscillator. The system possesses both kinetic energy and potential energy at this instant as shown in the bar graph. 444 CHAPTER 15 | Oscillatory Motion Figure 15.11 (a) If the atoms in a molecule U do not move too far from their equilibrium positions, a graph of potential energy versus separation distance between atoms is similar to the graph of potential energy versus posi- r tion for a simple harmonic oscillator (dashed black curve). (b) The forces between atoms in a solid can be modeled by imagining a b springs between neighboring atoms. Figure 15.11a shows that for small displacements from the equilibrium position, the potential energy curve for this function approximates a parabola, which represents the potential energy function for a simple harmonic oscillator. Therefore, we can model the complex atomic binding forces as being due to tiny springs as depicted in Figure 15.11b. The ideas presented in this chapter apply not only to block–spring systems and atoms, but also to a wide range of situations that include bungee jumping, playing a musical instrument, and viewing the light emitted by a laser. You will see more examples of simple harmonic oscillators as you work through this book. Ex a m pl e 15.3 Oscillations on a Horizontal Surface A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, horizontal air track. (A) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm. SOLUTION Conceptualize The system oscillates in exactly the same way as the block in Active Figure 15.10, so use that figure in your mental image of the motion. Categorize The cart is modeled as a particle in simple harmonic motion. Analyze Use Equation 15.21 to express the total energy E 5 12kA2 5 12mv 2max of the oscillator system and equate it to the kinetic energy of the system when the cart is at x 5 0: k 20.0 N/m Solve for the maximum speed and substitute numerical v max 5 A5 1 0.030 0 m 2 5 0.190 m/s Å m Å 0.500 kg values: (B) What is the velocity of the cart when the position is 2.00 cm? SOLUTION k 2 Use Equation 15.22 to evaluate the velocity: v56 1A 2 x22 Åm 20.0 N/m 56 3 1 0.030 0 m 2 2 2 1 0.020 0 m 2 2 4 Å 0.500 kg 5 60.141 m/s The positive and negative signs indicate that the cart could be moving to either the right or the left at this instant. (C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm. SOLUTION Use the result of part (B) to evaluate the kinetic energy K 5 12mv 2 5 12 1 0.500 kg 2 1 0.141 m/s 2 2 5 5.00 3 1023 J at x 5 0.020 0 m: Evaluate the elastic potential energy at x 5 0.020 0 m: U 5 12kx 2 5 12 1 20.0 N/m 2 1 0.0200 m 2 2 5 4.00 3 1023 J  15.4 | Comparing Simple Harmonic Motion with Uniform Circular Motion 445 15.3 cont. Finalize The sum of the kinetic and potential energies in part (C) is equal to the total energy, which can be found from Equation 15.21. That must be true for any position of the cart. WHAT IF? The cart in this example could have been set into motion by releasing the cart from rest at x 5 3.00 cm. What if the cart were released from the same position, but with an initial velocity of v 5 20.100 m/s? What are the new amplitude and maximum speed of the cart? Answer This question is of the same type we asked at the end of Example 15.1, but here we apply an energy approach. First calculate the total energy of the system at t 5 0: E 5 12mv 2 1 12kx 2 5 12 1 0.500 kg 2 1 20.100 m/s 2 2 1 12 1 20.0 N/m 2 1 0.030 0 m 2 2 5 1.15 3 1022 J Equate this total energy to the potential energy of the E 5 12kA2 system when the cart is at the endpoint of the motion: 2E 2 1 1.15 3 1022 J 2 Solve for the amplitude A: A5 5 5 0.033 9 m Å k Å 20.0 N/m Equate the total energy to the kinetic energy of the sys- E 5 12mv 2max tem when the cart is at the equilibrium position: 2E 2 1 1.15 3 1022 J 2 Solve for the maximum speed: v max 5 5 5 0.214 m/s Åm Å 0.500 kg The amplitude and maximum velocity are larger than the previous values because the cart was given an initial velocity at t 5 0. 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion Some common devices in everyday life exhibit a relationship between oscillatory motion and circular motion. For example, consider the drive mechanism for a non- electric sewing machine in Figure 15.12. The operator of the machine places her feet on the treadle and rocks them back and forth. This oscillatory motion causes the large wheel at the right to undergo circular motion. The red drive belt seen in the photograph transfers this circular motion to the sewing machine mechanism (above the photo) and eventually results in the oscillatory motion of the sewing The back edge of The oscillation of the treadle the treadle goes up causes circular motion of the and down as one’s drive wheel, eventually feet rock the treadle. resulting in additional up and down motion—of the sewing needle. John W. Jewett, Jr. Figure 15.12 The bottom of a treadle-style sewing machine from the early twentieth century. The treadle is the wide, flat foot pedal with the metal grillwork. 446 CHAPTER 15 | Oscillatory Motion needle. In this section, we explore this interesting relationship between these two The ball rotates like a particle in uniform types of motion. circular motion. Active Figure 15.13 is a view of an experimental arrangement that shows this relationship. A ball is attached to the rim of a turntable of radius A, which is illumi- nated from above by a lamp. The ball casts a shadow on a screen. As the turntable Lamp rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion. Consider a particle located at point P on the circumference of a circle of radius A as in Figure 15.14a, with the line OP making an angle f with the x axis at t 5 0. We call this circle a reference circle for comparing simple harmonic motion with uniform circular motion, and we choose the position of P at t 5 0 as our reference A position. If the particle moves along the circle with constant angular speed v until Turntable OP makes an angle u with the x axis as in Figure 15.14b, at some time t. 0 the angle A between OP and the x axis is u 5 vt 1 f. As the particle moves along the circle, the Screen projection of P on the x axis, labeled point Q , moves back and forth along the x axis between the limits x 5 6A. Notice that points P and Q always have the same x coordinate. From the right triangle OPQ , we see that this x coordinate is The ball’s shadow moves like a particle in simple x 1 t 2 5 A cos 1 vt 1 f 2 (15.23) harmonic motion. This expression is the same as Equation 15.6 and shows that the point Q moves with simple harmonic motion along the x axis. Therefore, simple harmonic motion ACTIVE FIGURE 15.13 along a straight line can be represented by the projection of uniform circular An experimental setup for demon- motion along a diameter of a reference circle. strating the connection between sim- This geometric interpretation shows that the time interval for one complete rev- ple harmonic motion and uniform circular motion. olution of the point P on the reference circle is equal to the period of motion T for simple harmonic motion between x 5 6A. Therefore, the angular speed v of P is the same as the angular frequency v of simple harmonic motion along the x axis (which is why we use the same symbol). The phase constant f for simple harmonic motion corresponds to the initial angle OP makes with the x axis. The radius A of the reference circle equals the amplitude of the simple harmonic motion. Because the relationship between linear and angular speed for circular motion is v 5 r v (see Eq. 10.10), the particle moving on the reference circle of radius A has a velocity of magnitude vA. From the geometry in Figure 15.14c, we see that the x component of this velocity is 2vA sin(vt 1 f). By definition, point Q has a velocity given by dx/dt. Differentiating Equation 15.23 with respect to time, we find that the velocity of Q is the same as the x component of the velocity of P. A particle is at At a later time t, the x The x component of The x component of the point P at t  0. coordinates of points P the velocity of P equals acceleration of P equals and Q are equal and are the velocity of Q. the acceleration of Q. given by Equation 15.23. y y y y S v v P P ax P vx P A t 0 A y S a f u x x x x O O x Q O vx Q O ax Q u  vt  f v  vA a  v 2A a b c d Figure 15.14 Relationship between the uniform circular motion of a point P and the simple harmonic motion of a point Q. A particle at P moves in a circle of radius A with constant angular speed v.  15.4 | Comparing Simple Harmonic Motion with Uniform Circular Motion 447 The acceleration of P on the reference circle is directed radially inward toward O and has a magnitude v 2/A 5 v2A. From the geometry in Figure 15.14d, we see Lamp that the x component of this acceleration is 2v2A cos(vt 1 f). This value is also the acceleration of the projected point Q along the x axis, as you can verify by taking the second derivative of Equation 15.23. Ball Turntable Quick Quiz 15.5 Figure 15.15 shows the position of an object in uniform cir- 0.50 m cular motion at t 5 0. A light shines from above and projects a shadow of the object on a screen below the circular motion. What are the correct values for the amplitude and phase constant (relative to an x axis to the right) of the Screen simple harmonic motion of the shadow? (a) 0.50 m and 0 (b) 1.00 m and 0 (c) 0.50 m and p (d) 1.00 m and p Figure 15.15 (Quick Quiz 15.5) An object moves in circular motion, casting a shadow on the screen below. Its position at an instant of time is shown. Ex a m pl e 15.4 Circular Motion with Constant Angular Speed The ball in Active Figure 15.13 rotates counterclockwise in a circle of radius 3.00 m with a constant angular speed of 8.00 rad/s. At t 5 0, its shadow has an x coordinate of 2.00 m and is moving to the right. (A) Determine the x coordinate of the shadow as a function of time in SI units. SOLUTION Conceptualize Be sure you understand the relationship between circular motion of the ball and simple harmonic motion of its shadow as described in Active Figure 15.13. Notice that the shadow is not at is maximum position at t 5 0. Categorize The ball on the turntable is a particle in uniform circular motion. The shadow is modeled as a particle in simple harmonic motion. Analyze Use Equation 15.23 to write an expression for x 5 A cos 1 vt 1 f 2 the x coordinate of the rotating ball: x Solve for the phase constant: f 5 cos21 a b 2 vt A 2.00 m Substitute numerical values for the initial conditions: f 5 cos21 a b 2 0 5 648.2° 5 60.841 rad 3.00 m If we were to take f 5 10.841 rad as our answer, the shadow would be moving to the left at t 5 0. Because the shadow is moving to the right at t 5 0, we must choose f 5 20.841 rad. Write the x coordinate as a function of time: x 5 3.00 cos (8.00t 2 0.841) (B) Find the x components of the shadow’s velocity and acceleration at any time t. SOLUTION dx Differentiate the x coordinate with respect to time to vx 5 5 1 23.00 m 2 1 8.00 rad/s 2 sin 1 8.00t 2 0.841 2 dt find the velocity at any time in m/s: 5 224.0 sin (8.00t 2 0.841) dv x Differentiate the velocity with respect to time to find the ax 5 5 1 224.0 m/s 2 1 8.00 rad/s 2 cos 1 8.00t 2 0.841 2 dt acceleration at any time in m/s2: 5 2192 cos (8.00t 2 0.841) Finalize These results are equally valid for the ball moving in uniform circular motion and the shadow moving in simple harmonic motion. Notice that the value of the phase constant puts the ball in the fourth quadrant of the xy coordinate system of Figure 15.14, which is consistent with the shadow having a positive value for x and moving toward the right. 448 CHAPTER 15 | Oscillatory Motion When u is small, a simple pendulum's motion can be 15.5 The Pendulum modeled as simple harmonic The simple pendulum is another mechanical system that exhibits periodic motion. motion about the equilibrium It consists of a particle-like bob of mass m suspended by a light string of length L position u  0. that is fixed at the upper end as shown in Active Figure 15.16. The motion occurs in the vertical plane and is driven by the gravitational force. We shall show that, provided the angle u is small (less than about 108), the motion is very close to that of a simple harmonic oscillator. S u The forces acting on the bob are the force T exerted by the string and the gravi- S S tational force mg. The tangential component mg sin u of the gravitational force T always acts toward u 5 0, opposite the displacement of the bob from the lowest posi- L tion. Therefore, the tangential component is a restoring force, and we can apply s m Newton’s second law for motion in the tangential direction: m g sin u d 2s u Ft 5 ma t S 2mg sin u 5 m m g cos u dt 2 S mg where the negative sign indicates that the tangential force acts toward the equi- librium (vertical) position and s is the bob’s position measured along the arc. We ACTIVE FIGURE 15.16 have expressed the tangential acceleration as the second derivative of the position A simple pendulum. s. Because s 5 Lu (Eq. 10.1a with r 5 L) and L is constant, this equation reduces to d 2u g 2 5 2 sin u dt L Considering u as the position, let us compare this equation with Equation 15.3. Does it have the same mathematical form? The right side is proportional to sin u rather than to u; hence, we would not expect simple harmonic motion because this expression is not of the same mathematical form as Equation 15.3. If we assume u is small (less than about 108 or 0.2 rad), however, we can use the small angle approximation, in which sin u < u, where u is measured in radians. Table 15.1 shows angles in degrees and radians and the sines of these angles. As long as u is less than Pitfall Prevention 15.5 approximately 108, the angle in radians and its sine are the same to within an accu- Not True Simple Harmonic Motion The pendulum does not exhibit racy of less than 1.0%. true simple harmonic motion for any Therefore, for small angles, the equation of motion becomes angle. If the angle is less than about 108, the motion is close to and can be d 2u g 2 5 2 u (for small values of u) (15.24) modeled as simple harmonic. dt L Equation 15.24 has the same mathematical form as Equation 15.3, so we conclude that the motion for small amplitudes of oscillation can be modeled as simple har- monic motion. Therefore, the solution of Equation 15.24 is u 5 umax cos(vt 1 f), where umax is the maximum angular position and the angular frequency v is g Angular frequency for X v5 (15.25) a simple pendulum ÅL The period of the motion is 2p L Period of a simple pendulum X T5 5 2p (15.26) v Åg In other words, the period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravity. Because the period is independent of the mass, we conclude that all simple pendula that are of equal length and are at the same location (so that g is constant) oscillate with the same period. The simple pendulum can be used as a timekeeper because its period depends only on its length and the local value of g. It is also a convenient device for making precise measurements of the free-fall acceleration. Such measurements are impor-  15.5 | The Pendulum 449 TABLE 15.1 Angles and Sines of Angles Angle in Degrees Angle in Radians Sine of Angle Percent Difference 08 0.000 0 0.000 0 0.0% 18 0.017 5 0.017 5 0.0% 28 0.034 9 0.034 9 0.0% 38 0.052 4 0.052 3 0.0% 58 0.087 3 0.087 2 0.1% 108 0.174 5 0.173 6 0.5% 158 0.261 8 0.258 8 1.2% 208 0.349 1 0.342 0 2.1% 308 0.523 6 0.500 0 4.7% tant because variations in local values of g can provide information on the location of oil and other valuable underground resources. Quick Quiz 15.6 A grandfather clock depends on the period of a pendulum to keep correct time. (i) Suppose a grandfather clock is calibrated correctly and then a mischievous child slides the bob of the pendulum downward on the oscillating rod. Does the grandfather clock run (a) slow, (b) fast, or (c) cor- rectly? (ii) Suppose a grandfather clock is calibrated correctly at sea level and is then taken to the top of a very tall mountain. Does the grandfather clock now run (a) slow, (b) fast, or (c) correctly? Ex a m pl e 15.5 A Connection Between Length and Time Christian Huygens (1629–1695), the greatest clockmaker in history, suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1 s. How much shorter would our length unit be if his suggestion had been followed? SOLUTION Conceptualize Imagine a pendulum that swings back and forth in exactly 1 second. Based on your experience in observ- ing swinging objects, can you make an estimate of the required length? Hang a small object from a string and simulate the 1-s pendulum. Categorize This example involves a simple pendulum, so we categorize it as an application of the concepts introduced in this section. T 2g 1 1.00 s 2 2 1 9.80 m/s2 2 Analyze Solve Equation 15.26 for the length and substi- L5 5 5 0.248 m 4p 2 4p 2 tute the known values: Finalize The meter’s length would be slightly less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g because the time has been defined to be exactly 1 s. WHAT IF? What if Huygens had been born on another planet? What would the value for g have to be on that planet such that the meter based on Huygens’s pendulum would have the same value as our meter? Answer Solve Equation 15.26 for g: 4p 2L 4p 2 1 1.00 m 2 g5 5 5 4p 2 m/s2 5 39.5 m/s2 T 2 1 1.00 s 2 2 No planet in our solar system has an acceleration due to gravity that large. Physical Pendulum Suppose you balance a wire coat hanger so that the hook is supported by your extended index finger. When you give the hanger a small angular displacement with your other hand and then release it, it oscillates. If a hanging object oscillates 450 CHAPTER 15 | Oscillatory Motion about a fixed axis that does not pass through its center of mass and the object can- not be approximated as a point mass, we cannot treat the system as a simple pendu- Pivot O lum. In this case, the system is called a physical pendulum. u Consider a rigid object pivoted at a point O that is a distance d from the center of d mass (Fig. 15.17). The gravitational force provides a torque about an axis through O, and the magnitude of that torque is mgd sin u, where u is as shown in Figure CM 15.17. We model the object as a rigid object under a net torque and use the rota- d sin u tional form of Newton’s second law, S text 5 Ia, where I is the moment of inertia of the object about the axis through O. The result is d 2u 2mgd sin u 5 I dt 2 S mg The negative sign indicates that the torque about O tends to decrease u. That is, the gravitational force produces a restoring torque. If we again assume u is small, the Figure 15.17 A physical pendulum approximation sin u < u is valid and the equation of motion reduces to pivoted at O. d 2u mgd 2 5 2a bu 5 2v 2u (15.27) dt I Because this equation is of the same mathematical form as Equation 15.3, its solu- tion is that of the simple harmonic oscillator. That is, the solution of Equation 15.27 is given by u 5 umax cos(vt 1 f), where umax is the maximum angular position and mgd v5 Å I The period is 2p I Period of a physical X T5 5 2p (15.28) pendulum v Å mgd This result can be used to measure the moment of inertia of a flat, rigid object. If the location of the center of mass—and hence the value of d—is known, the moment of inertia can be obtained by measuring the period. Finally, notice that Equation 15.28 reduces to the period of a simple pendulum (Eq. 15.26) when I 5 md 2, that is, when all the mass is concentrated at the center of mass. Ex a m pl e 15.6 A Swinging Rod A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane (Fig. 15.18). Find O the period of oscillation if the amplitude of the motion is Pivot small. d SOLUTION L Conceptualize Imagine a rod swinging back and forth when CM pivoted at one end. Try it with a meterstick or a scrap piece of wood. Categorize Because the rod is not a point particle, we cat- Figure 15.18 (Example 15.6) A rigid rod oscillating Mg S egorize it as a physical pendulum. about a pivot through one end is a physical pendulum Analyze In Chapter 10, we found that the moment of iner- with d 5 L/2. tia of a uniform rod about an axis through one end is 13ML2. The distance d from the pivot to the center of mass of the rod is L/2.  15.6 | Damped Oscillations 451 15.6 cont. 1 2 3 ML 2L Substitute these quantities into Equation 15.28: T 5 2p 5 2p Å Mg 1 L/2 2 Å 3g Finalize In one of the Moon landings, an astronaut walking on the Moon’s surface had a belt hanging from his space suit, and the belt oscillated as a physical pendulum. A scientist on the Earth observed this motion on television and used it to estimate the free-fall acceleration on the Moon. How did the scientist make this calculation? Torsional Pendulum Figure 15.19 shows a rigid object such as a disk suspended by a wire attached at the top to a fixed support. When the object is twisted through some angle u, the twisted wire exerts on the object a restoring torque that is proportional to the angular posi- tion. That is, t 5 2ku where k (Greek letter kappa) is called the torsion constant of the support wire and is a rotational analog to the force constant k for a spring. The value of k can be O obtained by applying a known torque to twist the wire through a measurable angle umax u. Applying Newton’s second law for rotational motion, we find that P d 2u o t 5 Ia S 2ku 5 I dt 2 The object oscillates about the 2 line OP with an amplitude umax. d u k 2 5 2 u (15.29) dt I Figure 15.19 A torsional pendulum. Again, this result is the equation of motion for a simple harmonic oscillator, with v 5 !k/I and a period I T 5 2p (15.30) W Period of a torsional Åk pendulum This system is called a torsional pendulum. There is no small-angle restriction in this situation as long as the elastic limit of the wire is not exceeded. 15.6 Damped Oscillations The oscillatory motions we have considered so far have been for ideal systems, that is, systems that oscillate indefinitely under the action of only one force, a linear restoring force. In many real systems, nonconservative forces such as friction or air resistance retard the motion. Consequently, the mechanical energy of the sys- tem diminishes in time, and the motion is said to be damped. The lost mechanical energy is transformed into internal energy in the object and the retarding medium. Figure 15.20 depicts one such system: an object attached to a spring and submersed in a viscous liquid. The opening photograph for this chapter depicts damped oscil- lations in practice. The piston-like devices below the sphere are dampers that trans- form mechanical energy of the oscillating sphere into internal energy. One common type of retarding force is that discussed in Section 6.4, where the force is proportional to the speed of the moving object and acts in the direc- tion opposite the velocity of the object with respect to the medium. This retarding force is often observed when an object S moves through air, for instance. Because m S the retarding force can be expressed as R 5 2b v (where b is a constant called the damping coefficient) and the restoring force of the system is 2kx, we can write New- ton’s second law as o Fx = 2kx 2 bvx = max Figure 15.20 One example of a damped oscillator is an object 2 dx d x attached to a spring and submersed 2kx 2 b 5m 2 (15.31) dt dt in a viscous liquid. 452 CHAPTER 15 | Oscillatory Motion x The solution to this equation requires mathematics that may be unfamiliar to you; The amplitude decreases as Ae (b/2m)t. we simply state it here without proof. When the retarding force is small compared A with the maximum restoring force—that is, when b is small—the solution to Equa- tion 15.31 is x 5 Ae2(b/2m)t cos (vt 1 f) (15.32) 0 t where the angular frequency of oscillation is k b 2 v5 2a b (15.33) Åm 2m This result can be verified by substituting Equation 15.32 into Equation 15.31. It ACTIVE FIGURE 15.21 is convenient to express the angular frequency of a damped oscillator in the form Graph of position versus time for a damped oscillator. b 2 v5 v 02 2 a b Å 2m where v 0 5 !k/m represents the angular frequency in the absence of a retarding force (th