Vibration and wave-solutions

# Vibration and wave-solutions - le(txl65 Vibration and wave...

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le (txl65) – Vibration and wave – mohanty – (81440) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A piston in an automobile engine is in simple harmonic motion. Its amplitude oF oscilla- tion From equilibrium (centered) position is ± 3 . 46 cm and its mass is 1 . 27 kg. ±ind the maximum velocity oF the piston when the auto engine is running at the rate oF 3470 rev / min. Correct answer: 12 . 5729 m / s. Explanation: The simple harmonic motion is described by x = A cos ω t, where ω is the Frequency in rad/s iF t is in seconds. To fnd ω From the given value in rev/min, Note: 60 rev / min = 1 rev / s = 1 Hz, so ω = 2 π f = 2 π (57 . 8334 rev / s) = 363 . 378 rad / s . Di²erentiating x above we fnd the velocity v = d x dt = - A ω sin( ω t ) meaning the maximum velocity is v max = A ω = (0 . 0346 m) (363 . 378 rad / s) = 12 . 5729 m / s , since sine has a maximum value oF 1. 002 10.0 points As your hand moves back and Forth to gener- ate longitudinal pulses in a spiral spring, your hand completes 2 . 13 back-and-Forth cycles ev- ery 5 . 82 s. The velocity oF the pulse in the spring is 1 . 03 cm / s. What is the wavelength? Correct answer: 0 . 0281437 m. Explanation: ±requence is the number oF cycles generated per unit time, so f = n t The velocity oF the wave is defned by v = f λ = n t λ λ = v t n = (1 . 03 cm / s) (5 . 82 s) 2 . 13 · 1 m 100 cm = 0 . 0281437 m . keywords: 003 (part 1 of 2) 10.0 points A sinusoidal wave oF the Form y = A sin( k x - ω t ) is traveling along a string in the x direc- tion, where A = 0 . 3 mm , k = 1 . 6 m 1 , ω = 34 rad / s , with x in meters and t in seconds. ±or this string, the mass per unit length is given by μ = 0 . 01 kg / m. 0 . 185 s 0 . 185 s 0 . 3 mm - 0 . 3 mm 3 . 93 m 3 . 93 m 0 . 3 mm - 0 . 3 mm Δ m What is the maximum value oF the acceler- ation oF the segment oF the string at x = 0 . 5 m?

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le (txl65) – Vibration and wave – mohanty – (81440) 2 Correct answer: 0 . 3468 m / s 2 . Explanation: Let : A = 0 . 3 mm = 0 . 0003 m and ω = 34 rad / s . The maximum acceleration is independent of x : a max = p d 2 y dt 2 P max = A ω 2 = (0 . 0003 m) (34 rad / s) 2 = 0 . 3468 m / s 2 . 004 (part 2 of 2) 10.0 points For a length segment Δ x = 1 cm along the string, what is the total energy of oscillation? Correct answer: 5 . 202 × 10 9 J. Explanation: Let : μ = 0 . 01 kg / m and Δ x = 1 cm = 0 . 01 m . The total energy is Δ E = Δ K + Δ U = Δ K max = 1 2 Δ m v 2 max = 1 2 μ Δ x ( ω A ) 2 = 1 2 (0 . 01 kg / m) (0 . 01 m) × (34 rad / s) 2 (0 . 0003 m) 2 = 5 . 202 × 10 9 J .
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Vibration and wave-solutions - le(txl65 Vibration and wave...

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