Measuring Simple Harmonic Motion

SECTION 2 Plan and Prepare Preview Vocabulary Scientific Meanings Explain that everyday words have more specialized meanings in science. Ask students the meaning of frequency. They are likely to use the word as a qualifier meaning how often something occurs. Explain than in science, frequency is used in a quantitative sense. It measures the number of cycles in a unit of time. Teach Demonstration Period and Frequency Purpose Show period and frequency empirically and verify their inverse relationship. Materials pendulum bob, string, ring stand, clock Procedure Attach the pendulum bob to the string, and suspend the string from the ring stand. Set the pendulum in motion. Have a student record the time required to complete 20 oscillations. Meanwhile, have another student determine how many oscillations take place each second. Have students use the first measurement to find the pendulum s period (T = number of seconds/20) and ask the students what the second measurement indicates. frequency Compare the values for the period and frequency of the pendulum. The two should be inversely related. Objectives Identify the amplitude of vibration. Recognize the relationship between period and frequency. Calculate the period and frequency of an object vibrating with simple harmonic motion. amplitude the maximum displacement from equilibrium period the time that it takes a complete cycle to occur frequency the number of cycles or vibrations per unit of time FIGURE 2.1 SECTION 2 Period and Frequency For any periodic motion such as the motion of this amusement park ride in Helsinki, Finland period and frequency are inversely related. 372 Chapter 11 Differentiated Instruction Below Level Students may be confused by the transition from cycles/s as a measure of frequency to the SI unit s 1 (hertz). Point out that the term cycle is not part of the SI unit because this term refers to an event rather than a unit of measure. The same holds true for the period, which can be considered as s/cycle but whose SI unit is simply s. This should help clarify the inverse relationship between frequency (cycles/s or s 1 ) and period (s/cycle or s). Measuring Simple Harmonic Motion Key Terms amplitude period frequency Amplitude, Period, and Frequency In the absence of friction, a moving trapeze always returns to the same maximum displacement after each swing. This maximum displacement from the equilibrium position is the amplitude. A pendulum s amplitude can be meas ured by the angle between the pendulum s equilibrium position and its maximum displacement. For a mass-spring system, the amplitude is the maximum amount the spring is stretched or compressed from its equilibrium position. Period and frequency measure time. Imagine the ride shown in Figure 2.1 swinging from maximum displacement on one side of equilibrium to maximum displacement on the other side, and then back again. This cycle is considered one complete cycle of motion. The period, T, is the time it takes for this complete cycle of motion. For example, if one complete cycle takes 20 s, then the period of this motion is 20 s. Note that after the time T, the object is back where it started. The number of complete cycles the ride swings through in a unit of time is the ride s frequency, f. If one complete cycle takes 20 s, then the ride s frequency is 1 cycles/s, or 0.05 cycles/s. The SI unit of frequency is 20 s 1, known as hertz (Hz). In this case, the ride s frequency is 0.05 Hz. Period and frequency can be confusing because both are concepts involving time in simple harmonic motion. Notice that the period is the time per cycle and that the frequency is the number of cycles per unit time, so they are inversely related. f = 1_ or T = 1_ f T This relationship was used to determine the frequency of the ride. f = 1_ = 0.05 Hz T = 1_ 20 s In any problem where you have a value for period or frequency, you can calculate the other value. These terms are summarized in Figure 2.2. Untitled-268 372 5/16/2011 7:06:45 Palais Didier/SuperStock 372 Chapter 11

Figure 2.2 Measures of simple HarMonic Motion Term Example Definition SI unit amplitude period, T frequency, f maximum displacement from equilibrium time that it takes to complete a full cycle number of cycles or vibrations per unit of time radian, rad meter, m second, s hertz, Hz (Hz = s 1 ) The period of a simple pendulum depends on pendulum length and free-fall acceleration. Although both a simple pendulum and a mass-spring system vibrate with simple harmonic motion, calculating the period and frequency of each requires a separate equation. This is because in each, the period and frequency depend on different physical factors. Consider an experimental setup of two pendulums of the same length but with bobs of different masses. The length of a pendulum is measured from the pivot point to the center of mass of the pendulum bob. If you were to pull each bob aside the same small distance and then release them at the same time, each pendulum would complete one vibration in the same amount of time. If you then changed the amplitude of one of the pendulums, you would find that they would still have the same period. Thus, for small amplitudes, the period of a pendulum does not depend on the mass or on the amplitude. However, changing the length of a pendulum does affect its period. A change in the free-fall acceleration also affects the period of a pendulum. The exact relationship between these variables can be derived mathematically or found experimentally. Did YOU Know? Galileo is credited as the first person to notice that the motion of a pendulum depends on its length and is independent of its amplitude (for small angles). He supposedly observed this while attending church services at a cathedral in Pisa. The pendulum he studied was a swinging chandelier that was set in motion when someone bumped it while lighting the candles. Galileo is said to have measured its frequency, and hence its period, by timing the swings with his pulse. Demonstration Relationship Between the Length and the Period of a Pendulum Purpose Verify the equation for a pendulum s period experimentally. Materials pendulum bob, string, ring stand, clock, meterstick Procedure Repeat the demonstration Period and Frequency with a variety of lengths. Record each length and its corresponding period. (Frequency does not need to be measured in this demonstration.) Verify that the results are consistent with the following equation: T = 2π Ç L a g Next ask the students to calculate the length required for a pendulum to have a period of 1.0 s. Have the students construct such a pendulum to test their prediction. Period of a Simple Pendulum in Simple Harmonic Motion T = 2π ÇÇL _ a g period = 2π square root of (length divided by free-fall acceleration) Problem Solving Vibrations and Waves 373 Reality Check titled-268 373 In order for a pendulum to approximate simple harmonic motion (and for this equation to apply), its amplitude must be less than about 15. For greater amplitudes, the pendulum s amplitude does affect its period. In many real-world cases, this equation does not apply However, it is still generally true that the shorter the length of the pendulum, the shorter the period. 5/16/2011 7:06:46 AM Vibrations and Waves 373

Teach continued Classroom Practice SHM of a Simple Pendulum What is the period of a 3.98 m long pendulum? What is the period of a 99.4 cm long pendulum? Answer: 4.00 s; 2.00 s A desktop toy pendulum swings back and forth once every 1.0 s. How long is this pendulum? Answer: 0.25 m What is the free-fall acceleration at a location where a 6.00 m long pendulum swings through exactly 100 cycles in 492 s? Answer: 9.79 m/s 2 PROBLEM guide b Use this guide to assign problems. SE = Student Edition Textbook PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) Solving for: L SE Sample, 1 3; Ch. Rvw. 19 PW 4, 5 PB 5 7 T, f SE 4; Ch. Rvw. 20, 27a PW Sample, 1 3 PB 8 10 a g SE 6 PB Sample, 1 4 *Challenging Problem FIGURE 2.3 Pendulums When the length of one pendulum is decreased, the distance that the pendulum travels to equilibrium is also decreased. Because the accelerations of the two pendulums are equal, the shorter pendulum will have a smaller period. L 1 L 2 m m Simple Harmonic Motion of a Simple Pendulum Why does the period of a pendulum depend on pendulum length and free-fall acceleration? When two pendulums have different lengths but the same amplitude, the shorter pendulum will have a smaller arc to travel through, as shown in Figure 2.3. Because the distance from maximum displacement to equilibrium is less while the acceleration caused by the restoring force remains the same, the shorter pendulum will have a shorter period. Why don t mass and amplitude affect the period of a pendulum? When the bobs of two pendulums differ in mass, the heavier mass provides a larger restoring force, but it also needs a larger force to achieve the same acceleration. This is similar to the situation for objects in free fall, which all have the same acceleration regardless of their mass. Because the acceleration of both pendulums is the same, the period for both is also the same. For small angles (less than 15 ), when the amplitude of a pendulum increases, the restoring force also increases proportionally. Because force isproportional to acceleration, the initial acceleration will be greater. However, the distance this pendulum must cover is also greater. For small angles, the effects of the two increasing quantities cancel and the pendulum s period remains the same. Sample Problem B You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling almost touches the floor and that its period is 12 s. How tall is the tower? ANALYZE Given: T = 12 s a g = g = 9.81 m/s 2 PLAN 374 Chapter 11 Problem Solving Unknown: L =? Use the equation for the period of a simple pendulum, and solve for L. T = 2π _ a L g T a _ g = L 2π T 2 a _ g 4π 2 = L SOLVE L = (12 s)2 (9.81 m/s 2 ) 4π 2 L = 36 m PREMIUM CONTENT Interactive Demo HMDScience.com Tips and Tricks Remember that on Earth s surface, a g = g = 9.81 m/s 2. Use this value in the equation for the period of a pendulum if a problem does not specify otherwise. At higher altitudes or on different planets, use the given value of a g instead. Continued Take It Further If a pendulum is at a higher elevation, the force of gravity is slightly less. If the building mentioned in the Sample Problem were at a significantly higher elevation, would this mean that the building was shorter or taller? Untitled-268 374 5/16/2011 7:06:47 L = (12s)2 _ (a g ) 4π 2 a g < g = 9.8 m/s 2 Therefore, the building is shorter. 374 Chapter 11

Robert Mathena/Fundamental Photographs, New York Simple Harmonic Motion of a Simple Pendulum (continued) 1. If the period of the pendulum in the preceding sample problem were 24 s, how tall would the tower be? 2. You are designing a pendulum clock to have a period of 1.0 s. How long should the pendulum be? 3. A trapeze artist swings in simple harmonic motion with a period of 3.8 s. Calculate the length of the cables supporting the trapeze. 4. Calculate the period and frequency of a 3.500 m long pendulum at the following locations: a. the North Pole, where a g = 9.832 m/s 2 b. Chicago, where a g = 9.803 m/s 2 c. Jakarta, Indonesia, where a g = 9.782 m/s 2 Period of a mass-spring system depends on mass and spring constant. Now consider the period of a mass-spring system. In this case, according to Hooke s law, the restoring force acting on the mass is determined by the displacement of the mass and by the spring constant (F elastic = kx). The magnitude of the mass does not affect the restoring force. So, unlike in the case of the pendulum, in which a heavier mass increased both the force on the bob and the bob s inertia, a heavier mass attached to a spring increases inertia without providing a compensating increase in force. Because of this increase in inertia, a heavy mass has a smaller acceleration than a light mass has. Thus, a heavy mass will take more time to complete one cycle of motion. In other words, the heavy mass has a greater period. Thus, as mass increases, the period of vibration increases when there is no compensating increase in force. Conceptual Challenge Pendulum on the Moon The free-fall acceleration on the surface of the moon is approximately one-sixth of the free-fall acceleration on the surface of Earth. Compare the period of a pendulum on the moon with that of an identical pendulum set in motion on Earth. Pendulum Clocks Why is a wound mainspring often used to provide energy to a pendulum clock in order to prevent the amplitude of the pendulum from decreasing? Answers Practice B 1. 1.4 10 2 m 2. 25 cm 3. 3.6 m 4. a. 3.749 s, 0.2667 Hz b. 3.754 s, 0.2664 Hz c. 3.758 s, 0.2661 Hz Teaching Tip Previously, students compared a simple pendulum with a mass-spring system to find their similarities (see Figure 1.6). The discussion of mass on this page can be used to demonstrate their differences. Mass does not affect the period of a pendulum, but it does affect the period of a mass-spring system. This difference exists because the restoring forces are different. Answers Conceptual Challenge 1. The period of the pendulum on the moon would be a little less than 2.5 times as long as the period on Earth. 2. Because a pendulum s motion only approximates simple harmonic motion, if the amplitude decreased, the clock would not keep accurate time. Vibrations and Waves 375 led-268 375 Deconstructing Problems Because the period in Problem B.1 is twice that of Sample Problem B, students may think that the height is also double. This will give them an incorrect answer of 72 m. Point out to students that the length increases as the square of the period. So if the period doubles, the length would increase by a factor of 2 2, or 4. The answer would be 4 36 m, or 144 m (1.4 10 2 m when rounded to two significant figures). 5/16/2011 7:06:48 AM Vibrations and Waves 375

Teach continued Classroom Practice SHM of a Mass-Spring System A 1.0 kg mass attached to one end of a spring completes one oscillation every 2.0 s. Find the spring constant. Answer: 9.9 N/m The greater the spring constant (k), the stiffer the spring; hence a greater force is required to stretch or compress the spring. When force is greater, acceleration is greater and the amount of time required for a single cycle should decrease (assuming that the amplitude remains constant). Thus, for a given amplitude, a stiffer spring will take less time to complete one cycle of motion than one that is less stiff. As with the pendulum, the equation for the period of a mass-spring system can be derived mathematically or found experimentally. Period of a Mass-Spring System in Simple Harmonic Motion T = 2π m_ k period = 2π square root of (mass divided by spring constant) Note that changing the amplitude of the vibration does not affect the period. This statement is true only for systems and circumstances in which the spring obeys Hooke s law. Simple Harmonic Motion of a Mass-Spring System PREMIUM CONTENT Interactive Demo HMDScience.com Sample Problem C The body of a 1275 kg car is supported on a frame by four springs. Two people riding in the car have a combined mass of 153 kg. When driven over a pothole in the road, the frame vibrates with a period of 0.840 s. For the first few seconds, the vibration approximates simple harmonic motion. Find the spring constant of a single spring. ANALYZE (1275 kg + 153 kg) Given: m = = 357 kg T = 0.840 s 4 Unknown: k =? Use the equation for the period of a mass-spring system, and solve for k. SOLVE T = 2π _ m k T 2 = 4π 2 ( m_ k ) k = _ 4π2 m T 2 = 4π2 (357 kg) (0.840 s) 2 k = 2.00 10 4 N/m 376 Chapter 11 Problem Solving Continued Alternative Approaches Think of the total mass (1275 kg + 153 kg = 1428 kg) as being supported by one spring with four times the strength of each spring: Untitled-268 376 5/16/2011 7:06:49 k = _ 4π2 m T 2 k = 4π2 (1428 kg) (0.840 s) 2 k = 7.99 10 4 N/m The spring constant of a single spring is one-fourth of this value. k = 0.25(7.99 10 4 N/m) k = 2.00 10 4 N/m 376 Chapter 11

Simple Harmonic Motion of a Mass-Spring System (continued) PROBLEM guide C 1. A mass of 0.30 kg is attached to a spring and is set into vibration with a period of 0.24 s. What is the spring constant of the spring? Use this guide to assign problems. SE = Student Edition Textbook 2. When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. What is the spring constant of the spring? PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) 3. A 125 N object vibrates with a period of 3.56 s when hanging from a spring. What is the spring constant of the spring? 4. When two more people get into the car described in Sample Problem C, the total mass of all four occupants of the car becomes 255 kg. Now what is the period of vibration of the car when it is driven over a pothole in the road? Solving for: k SE Sample, 1 3; PW 4 PB 4 6 5. A spring of spring constant 30.0 N/m is attached to different masses, and the system is set in motion. Find the period and frequency of vibration for masses of the following magnitudes: a. 2.3 kg b. 15 g c. 1.9 kg T, f SE 4,5; ch. Rvw.21 PW Sample, 1, 2, 3* PB 7 10 m PW 5,6 PB Sample, 1 3 *Challenging Problem SECTION 2 FORMATIVE ASSESSMENT Answers Reviewing Main Ideas 1. The reading on a metronome indicates the number of oscillations per minute. What are the frequency and period of the metronome s vibration when the metronome is set at 180? 2. A child swings on a playground swing with a 2.5 m long chain. a. What is the period of the child s motion? b. What is the frequency of vibration? 3. A 0.75 kg mass attached to a vertical spring stretches the spring 0.30 m. a. What is the spring constant? b. The mass-spring system is now placed on a horizontal surface and set vibrating. What is the period of the vibration? Practice C 1. 2.1 10 2 N/m 2. 25 N/m 3. 39.7 N/m 4. 0.869 s 5. a. 1.7 s, 0.59 Hz b. 0.14 s, 7.1 Hz c. 1.6 s, 0.62 Hz Critical Thinking 4. Two mass-spring systems vibrate with simple harmonic motion. If the spring constants are equal and the mass of one system is twice that of the other, which system has a greater period? Assess and Reteach Assess Use the Formative Assessment on this page to evaluate student mastery of the section. led-268 377 Answers to Section Assessment 1. 3.0 Hz, 0.33 s 2. a. 3.2 s b. 0.31 Hz 3. a. 25 N/m b. 1.1 s Vibrations and Waves 377 5/16/2011 7:06:49 AM Reteach For students who need additional instruction, download the Section Study Guide. Response to Intervention To reassess students mastery, use the Section Quiz, available to print or to take directly online at HMDScience.com. 4. The system with the larger mass has a greater period. Vibrations and Waves 377