Chapter 11 Simple Harmonic Motion In This Chapter Using Hooke's law Following simple harmonic motion Calculating simple harmonic motion velocity Finding simple harmonic motion acceleration Calculating a spring's period of oscillation Simple harmonic motion - the motion of springs - is a very important topic in physics. This kind of motion is all based on Hooke's law, which says that the force on a spring in simple harmonic motion is proportional to the distance the object is away from its equilibrium position (the location where the oscillator will feel no unbalanced force). In other words, the farther a spring is stretched, the more it pulls back. Hooking into Hooke's Law Hooke's law is represented by this equation: F = -kAx MBER This deceptively simple equation is at the heart of explaining the motion of objects on springs. It says that the force on an object in simple harmonic motion is proportional to the displacement (that's Ax) from rest. The constant of proportionality is & (Hooke's constant, also called a spring constant), which must be measured for every situation (because no two springs are exactly identical, for example). The negative sign In the equation Indicates that & is a restoring force - that Is, that the force points toward the equilibrium position of the object. What are the units of spring constants? Just check out the equation: Because F= -kAx, k must have the unit N/m. You'll commonly see Hooke's law applied to springs. Don't get confused by the minus sign In this equation; It's just there to indicate that the force opposes the displacement, which you know is true about springs.94 Part III: Being Energetic: Work TIP If it's easier for you to understand, put in the minus sign after you've done the rest of O the problem. You can easily figure out which way the force is going, and if it's in a positive direction as defined by the axes in the problem, the force is positive. SAMPLE 2. You're stretching a spring with A. The correct answer is -10.0 N. spring constant 5.0 N/m. If you 1. Use the equation F = -kAx. stretch it +2.0 m, what pull do you feel from the spring? 2. Plug in the numbers: . . F = -kAx = -(5.0 N/m)(2.0 m) =-10.0N 1. You have a spring whose spring constant is 2. You have a spring of spring constant 200 N/m, and you want to stretch it by 6.0 m. 50.0 N/m. What force do you need to What force do you need to apply? stretch it by 5.3 m? Salverit Solve It You have a spring of spring constant 4. 73 N/m. How much does the spring stretch It takes 200.0 N to stretch a spring 2.0 m. if you apply a force of 52 N? What is its spring constant? Solve It Solve JeChapter 11: Simple Harmonic Motion 195 Simply Simple Harmonic Motion Take a look at the spring in Figure 11-1. It starts at rest, drops down to position -A, and then moves back up to position A. Equilibrium -A Figure 11-1: A weight on a spring. @John Way & Sons, the. How do you describe this motion? In terms of position A, called the amplitude of the spring's motion: y = A cos (at) In this equation for displacement, / is time, and @ is the angular speed. (It's called angular speed here for a variety of reasons; it turns out that simple harmonic motion is only one component - hence the cosine - of full circular motion.) That equation assumes that you start at full extension, at time / = 0. If you start at a different time, (, you can adjust the equation to match the spring's motion like this: y = A cos [o(t-t)]196 Part III: Being Energetic: Work A weight on a spring is making that spring oscillate up and down. If the amplitude of the motion is 1.0 m, the angular speed is 1.0 radian/s, and the weight is Initially at its maximum upward position, where will the oscillating weight be after 10.0 seconds? A. The correct answer is -0.84 m. 1. Use the equation y = A cos (wl). 2. Plug In the numbers: y = A cos (of) = (1.0 m) cos [(1.0 radian/s) (10.0 s)] = -0.84 m The negative sign indicates that the spring Is extended downward. If you plug the numbers in this equation Into your calculator, either put it into radian mode O because of Is In radians or convert to degrees by multiplying ot by 180/m. 5. A spring with a weight on it has an ampli- 6. A spring with a weight pulling it down has tude of motion of 2.5 m and an angular an amplitude of motion of 5.0 m and an speed of 2.0 radians/s. If the weight starts angular speed of 16.0 radians/s. If the at maximum extension, where will it be weight starts at maximum extension, where after 60.0 seconds? will it be after 60.0 seconds? Solve It Solve It 7. A spring starts at maximum extension, is at -3.0 m at time 60.0 s, and has an angular 8. A spring starts at maximum extension, is speed of 6.0 radians/s. What is its at -4.5 m at time 10.0 s, and has an angular amplitude? speed of 16.0 radians/s. What is its amplitude? Solve it Solve leChapter 11: Simple Harmonic Motion 197 Getting Periodic A weight on the end of a spring bounces up and down periodically. The time it takes to bounce up and down, completing a full cycle and coming back to where it started, is called Its period, represented by symbol 7. Because y = A cos (@), the object goes through 2/r radians in period T, so you have this relation: In other words: The period, T, is measured in seconds. Besides the period, oscillations are measured in frequency, which is the number of cycles per second. The frequency of a simple harmonic oscillator is equal to the inverse of the period. The equation for frequency is Because a = 27 / 7, you can change the frequency equation to get this equation: = 21 = 2xf Frequency is measured in Hertz, abbreviation Hz, which is one cycle per second. ANICAL STUFF It's worth noting that you can call a angular speed, but when you're working with simple harmonic motion, it's usually called angular frequency. SAMPLE e. A weight on a spring is bouncing up and down with an angular frequency of A. The correct answer is 0.52 s. 12.0 radians/s. How long does it take to 1. Use this equation: complete each cycle? T = 21 2. Plug In the numbers: T = 21 = 12.0 radians/s - =0.52 s198 Part III: Being Energetic: Work 9. A spring has a weight with an angular fre- 70. A spring has a weight pulling It down, quency of 4.5 radians/s. What is the and the angular frequency of oscillation is weight's period? 1.5 radians/s. What is the weight's period? Solve it Solve It 1 1. A spring with a weight on It has an angular 12. A spring with a weight on it is bouncing up frequency of 0.70 radians/s. What is its and down with an angular frequency of frequency? 1.3 radians/s. What is Its frequency? Solve It Solve It 149 40 Considering Velocity As I explain in the earlier section "Simply Simple Harmonic Motion," the displace- ment of an object on a spring looks like this: y = A cos (wl) It turns out that you also can express the object's velocity; here's that equation: ", = -do sin 0 = -Aw sin (of) This equation assumes that you start at full extension, at time / = 0, which means the Initial velocity of the object is zero. If you want to adjust this equation to start at some other time, (, you alter the equation like so: U, = -Au sin [w(1 - ( )] When the object is at its fullest extension, the velocity Is zero, and when it's swooping back through the equilibrium point, Its velocity is at its maximum.Chapter 11: Simple Harmonic Motion 199 MPLE A weight on a spring is bouncing up and down with an angular frequency of 34 radians/s and an amplitude of 1.4 m. If the weight starts at maximum displacement at / = 0, what Is its speed at / - 5.0 67 A. The correct answer Is 4.6 m/s. 1. Use this equation: U, = -An sin (of) 2. Plug In the numbers: v, =-Au sin (of) =-(1.4 m)(3.4 radians/s) sin [(3.4 radians/s)(5.0 =) ] = 4.6 m/s 13. If you have a spring with a weight on it, 14. A spring with a weight on it is bouncing up and the weight's angular frequency is and down with an angular frequency of 1.7 radians/s and amplitude is 5.6 m, what 2.7 radians/s and amplitude of 1.7 m. What is the weight's velocity at 30.0 s? Is Its velocity at 10.0 s? Solve 16 Solve it Figuring the Acceleration In addition to calculating the displacement and velocity of an object on a spring, you can calculate its acceleration. Displacement goes from -4 to A, where A Is the amplitude. And the velocity of an object on a spring goes from -An to Aw; the speed is at Its minimum (zero) at elther end of the spring's maximum extension, and It's at a maximum at the equilibrium point (where the object would be at rest). The acceleration varies from -Aa to Aar, and It's at Its maximum when the object is speeding up (or down!) the most, which is at the ends of its oscillations (when the velocity is zero). The acceleration of the object is zero when the object is passing through the equilibrium position because the net force on the object is zero at the equilibrium position.200 Part Ill: Being Energetic: Work EMBER Here's the equation for the acceleration of an object on a spring: a = -An' cos 8 = -And cos (of) A weight on a spring is bouncing up and down with an angular frequency of 3.4 radians/s and an amplitude of 1.4 m. If the weight starts at maximum displacement, what Is its accel- eration at 5.0 s? A. The correct answer is 4.5 m/s?. 1. Use this equation: a=-Aor cos (of) 2. Plug in the numbers: a =-Am' cos (at) =-(1.4 m)(3.4 radians/s)" cos [(3.4 radians/s) (5.0 s) ] 15. A spring has a weight on it. If the weight's angular frequency is 1.3 radians/s and its 16. You have a spring with a weight pulling amplitude is 1.0 m, what is the weight's it down; the angular frequency is 1.7 radians/s and the amplitude is 6.0 m. acceleration at f = 4.9 s? What is the weight's acceleration at 60 s? Solve It Solve it 17. A spring with a weight on it has an angular 18. A spring with a weight on it is moving up frequency of 3.7 radians/s and an ampli- and down with an angular frequency of tude of 1.4 m. What is the weight's 1.3 radians/s and an amplitude of 2.9 m. acceleration at 15 s? What Is Its acceleration at 7.0 s? Solve It Solve ItChapter 11: Simple Harmonic Motion 20 Bouncing Around with Springs What is the period of a spring in terms of Its spring constant, k? You'll often come across that question in physics problems. You know that: F= -kAx You also know that F = ma, so: F = ma = -kAx For simple harmonic motion, you know that: x = A cos (wf) a --Aar cos (al) Putting this all together gives you: ma = -marcos (of)= -kAx = -kA cos (of) That lengthy equation becomes moz = k Which, if you solve for a, gives you the angular frequency of an object on a spring: @ =. Vm A weight on a spring is bouncing up A. The correct answer is y = A cos (al) = and down. The spring constant is 1.6 (3.0 m) cos (1.3 radians/s)t. N/m, the mass Is 1.0 kg, and the amplitude is 3.0 m. What equation 1. Use this equation: describes the spring's motion? _ 1.6 N/m = 1.3 radians/s V 1.0 kg 2. So this equation describes the motion: y = A cos (al) = (3.0 m) cos (1.3 radians/s)t202 Part Ill: Being Energetic: Work 19. If you have a spring with a mass of. 1.0 kg 20. What is the period of oscillation of a spring on it and a spring constant of 12.0 N/m, with a mass of 300 g on it and a spring what is the spring's period of oscillation? constant of 7.0 N/m? Solve it Solve It 27. A weight on a spring is moving up and 22. A weight on a spring is oscillating up and down. The spring constant is 1.9 N/m, the down. The spring constant is 2.3 N/m, the mass is 1.0 kg, and the amplitude is 2.4 m. mass is 1.3 kg, and the amplitude is 5.6 m. What equation describes the spring's What equation describes the spring's motion? motion? Solve It Solve JU Talking about Energy When you stretch a spring, you transfer potential energy to that spring (just like when you lift a weight against the force of gravity). For a spring of spring constant * stretched a distance x from equilibrium, the potential energy in the spring is PE = - kx' JMAPLE What is the potential energy In a spring 2. Plug In the numbers: of spring constant 40 N/m that's stretched 5.0 m from equilibrium? PE = kx A. The correct answer is 500 J. = =(40 N/m)(5.0 m)' 1. Use this equation: = 500 J PE = - kx'Chapter 11: Simple Harmonic Motion 203 23. If you stretch a spring of spring constant 24. If you stretch a spring of spring constant 100 N/m by 5.0 m, what potential energy 250 N/m by 6.5 m, what potential energy Is In the spring? Is in the spring? Solve It Following the Ticktock of Pendulums In addition to springs, physics problems about simple harmonic motion may ask you to deal with pendulums like the one in Figure 11-2. Figure 11-2 A pendulum. S John Wiley & Song, Inc. You can calculate the angular frequency of a pendulum of length L using the following equation, where g is the acceleration due to gravity: What's the angular frequency of a A. The correct answer Is 3.1 radians/s. pendulum of length 1.0 m? Use this equation and plug in the numbers: a= = (9.8 m/s =3.1 radians/s