# 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 heart’s 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 , ω = 2 π (5 Hz) = 10 π s − 1 and φ = 0 . Then x ( t ) = (10 cm) × cos[(10 π s − 1 ) t ] . (b) The maximum (positive) velocity occurs at t = 0 if sin φ = − 1 (from Equation 15-10), therefore, the motion is described by Equation 15-9 with A = 2 . 5 cm , ω = 5 s − 1 , and φ = − π/ 2 . Since cos( ωt − π/ 2) = sin ωt, 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 π 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 , ω ≃ 2 π/ 4 s ≃ 1 2 π s − 1 , and φ ≃ 0; (b) A ≃ 30 cm , ω ≃ 2 π/ 3 . 2 s ≃ 2 s − 1 , and φ ≃ − 90 ◦ ≃ − 1 2 π ; (c) A ≃ 40 cm , ω ≃ 2 π/ (2 × 2 s) ≃ 1 2 π s − 1 , and φ ≃ cos − 1 (27 / 40) ≃ 48 ◦ ≃ 1 4 π. 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 ω = radicalbig k/m = radicalbig (5 . 6 N / m) / (0 . 2 kg) = 5 . 29 s − 1 . Then (a) f = ω/ 2 π = 0 . 842 Hz (Equation 15-8b), (b) T = 1 /f = 1 . 19 s (Equation 15-8c), (c) v max =...
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## This note was uploaded on 05/18/2011 for the course UNKNOWN foud taught by Professor Asq during the Spring '11 term at Art Institute of Atlanta.

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

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