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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 151 and 152: 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 151). 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 151). 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 159 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 1510), therefore, the motion is described by Equation 159 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. 1533. Solution The amplitude is the maximum displacement, read along the x axis (ordinate) in Fig. 1533. 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 1533 Problem 5 Solution. Problem 6. A 200g 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 158a 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 158b), (b) T = 1 /f = 1 . 19 s (Equation 158c), (c) v max =...
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