chap15 - PART 2 OSCILLATIONS, WAVES, AND FLUIDS CHAPTER 15...

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Unformatted text preview: PART 2 OSCILLATIONS, WAVES, AND FLUIDS CHAPTER 15 OSCILLATORY MOTION ActivPhysics can help with these problems: All Activities in Section 9 Sections 15-1 and 15-2: Oscillations and Simple Harmonic Motion Problem 1. A doctor counts 77 heartbeats in one minute. What are the period and frequency of the hearts oscillations? Solution If 77 heartbeats take 1 min., then one heartbeat (one cycle) takes T = 1 min / 77 = 60 s / 77 = 0 . 77 s , which is the period. The frequency is f = 77 / min = 77 / 60 s = 1 . 28 Hz . (One can see that T = 1 /f. ) Problem 2. A violin string playing the note A oscillates at 440 Hz. What is the period of its oscillation? Solution T = 1 /f = 1 / 440 Hz = 2 . 27 ms (Equation 15-1). Problem 3. The vibration frequency of a hydrogen chloride molecule is 8 . 66 10 13 Hz . How long does it take the molecule to complete one oscillation? Solution T = 1 /f = 1 / (8 . 66 10 13 Hz) = 1 . 15 10 14 s = 11 . 5 fs (Equation 15-1). Problem 4. Write expressions for simple harmonic motion (a) with amplitude 10 cm, frequency 5.0 Hz, and with maximum displacement at t = 0 and (b) with amplitude 2.5 cm, angular frequency 5 . 0 s 1 , and with maximum velocity at t = 0 . Solution (a) Use Equation 15-9 with A = 10 cm , = 2 (5 Hz) = 10 s 1 and = 0 . Then x ( t ) = (10 cm) cos[(10 s 1 ) t ] . (b) The maximum (positive) velocity occurs at t = 0 if sin = 1 (from Equation 15-10), therefore, the motion is described by Equation 15-9 with A = 2 . 5 cm , = 5 s 1 , and = / 2 . Since cos( t / 2) = sin t, the expression simplifies to x ( t ) = (2 . 5 cm)sin[(5 s 1 ) t ] . Problem 5. Determine the amplitude, angular frequency, and phase constant for each of the simple harmonic motions shown in Fig. 15-33. Solution The amplitude is the maximum displacement, read along the x axis (ordinate) in Fig. 15-33. The angular frequency is 2 times the reciprocal of the period, which is the time interval between corresponding points read along the t axis (abscissa). The phase constant can be determined from the intercept and slope (displacement and velocity) at t = 0 . One sees that (a) A 20 cm , 2 / 4 s 1 2 s 1 , and 0; (b) A 30 cm , 2 / 3 . 2 s 2 s 1 , and 90 1 2 ; (c) A 40 cm , 2 / (2 2 s) 1 2 s 1 , and cos 1 (27 / 40) 48 1 4 . figure 15-33 Problem 5 Solution. Problem 6. A 200-g mass is attached to a spring of constant k = 5 . 6 N / m and set into oscillation with amplitude A = 25 cm . Determine (a) the frequency in hertz, (b) the period, (c) the maximum velocity, and (d) the maximum force in the spring. 232 CHAPTER 15 Solution For a mass on a spring, Equation 15-8a gives = radicalbig k/m = radicalbig (5 . 6 N / m) / (0 . 2 kg) = 5 . 29 s 1 . Then (a) f = / 2 = 0 . 842 Hz (Equation 15-8b), (b) T = 1 /f = 1 . 19 s (Equation 15-8c), (c) v max =...
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chap15 - PART 2 OSCILLATIONS, WAVES, AND FLUIDS CHAPTER 15...

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