HW12-solutions

# HW12-solutions - markowitz(am45362 – HW12 – distler...

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Unformatted text preview: markowitz (am45362) – HW12 – distler – (56295) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points At t = 0 , a transverse wave pulse in a wire is described by the function y ( x,t = 0) = 6 x 2 + 3 , where x and y are in meters. The pulse is traveling in the positive x direction with a speed of 4 . 5 m / s . Which formula correctly describes the pulse at time t ? 1. y = 4 . 5 t + 6 x 2 + 3 2. y = 4 . 5 t + 6 x 2 + 3 3. y = 6 ( x − 4 . 5 t ) 2 + 3 correct 4. y = 6 − 4 . 5 t x 2 + 3 5. y = 6 ( x + 4 . 5 t ) 2 + 3 6. y = − 4 . 5 t + 6 x 2 + 3 Explanation: For the wave traveling in the positive di- rection with the speed of 4 . 5 m / s, we need to replace x with x − 4 . 5 t which gives y = 6 ( x − 4 . 5 t ) 2 + 3 . keywords: 002 10.0 points When a particular wire is vibrating with a fre- quency of 3 . 6 Hz, a transverse wave of wave- length 49 . 8 cm is produced. Determine the speed of wave pulses along the wire. Correct answer: 1 . 7928 m / s. Explanation: Let : f = 3 . 6 Hz and λ = 49 . 8 cm . If the frequency is f and the wavelength is λ then the speed of wave is v = f λ = (3 . 6 Hz) (49 . 8 cm) · 1 m 100 cm = 1 . 7928 m / s . keywords: 003 10.0 points An object of mass 0.1 kg is attached to a spring of negligible mass and is executing sim- ple harmonic oscillation with an angular fre- quency of 3 rad/s. If v x (0) = +0 . 5 m/s and x (0) = − . 1 m, what is the phase φ ? The general solution for such an oscillator is x ( t ) = A cos( ω t + φ ) . 1. . 9911 2. − 1 . 039 3. − 3 . 142 4. 3 . 142 5. 4 . 173 6. − 2 . 112 correct 7. . 519 8. +2 . 112 9. Zero 10. 1 . 03 Explanation: x (0) = A cos φ and v x = − Aω sin( ω t + φ ) , so v x (0) = − Aω sin φ and markowitz (am45362) – HW12 – distler – (56295) 2 − Aω sin φ A cos φ = v x (0) x (0) tan φ = − v x (0) ω x (0) φ = tan − 1 bracketleftbigg − . 5 m / s (3 1 / s)( − . 1 m) bracketrightbigg = 1 . 03 according to your calcuator, which is incorrect because your calculator does not know any physics. In this specific case, x (0) ∝ cos φ < and v x (0) ∝ − sin φ > 0 (sin φ < 0) so φ = 1 . 03 − π = − 2 . 112 . 004 10.0 points One end of a string 2 . 56 m long is moved up and down with simple harmonic motion at a frequency of 69 Hz . The waves reach the other end of the string in 0 . 5 s . Find the wavelength of the waves on the string. Correct answer: 7 . 42029 cm. Explanation: Let : Δ x = 2 . 56 m , f = 69 Hz , and Δ t = 0 . 5 s . The wave velocity is v = Δ x Δ t λ = v f = Δ x f Δ t = 2 . 56 m (69 Hz) (0 . 5 s) = 7 . 42029 cm ....
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HW12-solutions - markowitz(am45362 – HW12 – distler...

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