Old Midterm 4 - agnew (ja26385) oldmidterm 04 Turner...

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agnew (ja26385) – oldmidterm 04 – Turner – (58220) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A particle oscillates harmonically x = A sin( ωt + φ 0 ) , with amplitude 8 m, angular Frequency π s 1 , and initial phase π 3 radians. Every now and then, the particle’s kinetic energy and poten- tial energy happen to be equal to each other ( K = U ). When does this equality happen For the frst time aFter t = 0? 1. 0.4167 s correct 2. 0.5267 s 3. 0.9967 s 4. 0.7615 s 5. 0.5884 s 6. 0.3467 s 7. 0.6547 s 8. 0.2238 s 9. 0.1294 s 10. 0.8623 s Explanation: Let : A = 8 m , ω = π s 1 , and φ 0 = π 3 . x = A cos( ω t + φ 0 ) v = sin( ω t + φ 0 ) and ω = r k m . K = U 1 2 mv 2 = 1 2 k x 2 A 2 ω 2 sin 2 ( ω t + φ 0 ) = k m A 2 cos 2 ( ω t + φ 0 ) ω R sin 2 ( ω t + φ 0 ) = r k m R cos 2 ( ω t + φ 0 ) v v v sin p π t + π 3 Pv v v = v v v cos p π t + π 3 Pv v v . Since | sin x | = | cos x | when x = (2 n + 1) π 4 , n = 0 , ± 1 , ± 2 ,..., (2 n + 1) π 4 = π t + π 3 2 n + 1 4 = t + 1 3 t = 2 n + 1 4 1 3 . Since t > 0 For n 1, the frst positive time is when t = 1: t = 0 . 416667 002 10.0 points Longitudinal sound waves cannot propagate through 1. steam 2. a vacuum correct 3. wet sand 4. solids 5. liquids 6. gases Explanation: A longitudinal sound wave is a pressure wave which can propagate through any media
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agnew (ja26385) – oldmidterm 04 – Turner – (58220) 2 whose pressure may vary, including gases, liq- uids, solids or even mixtures such as wet sand or steam (tiny water droplets in air / water vapor mixture). The one ‘medium’ through which a pressure wave cannot propagate is a vacuum . 003 10.0 points At t = 0, a transverse wave pulse in a wire is described by the function y = 5 e x 2 / 2 , where x and t are given in SI units. Which function describes this wave if it is traveling in the negative x direction with a speed of 4 . 5 m / s? 1. y = 5 e ( x 4 . 5 t ) 2 / 2 2. y = 5 e x 2 / 2 4 . 5 t 3. y = 5 e x 2 / 2 + 4 . 5 t 4. y = 5 e ( x 2 +4 . 5 t ) / 2 5. y = 5 e ( x +4 . 5 t ) 2 / 2 correct 6. y = 5 e x 2 / 2 7. y = 5 e ( x 2 4 . 5 t ) / 2 Explanation: For the wave traveling in the negative di- rection with the speed of 4 . 5 m / s , we need to replace x with x + 4 . 5 t which gives y = 5 e ( x +4 . 5 t ) 2 / 2 . keywords: 004 10.0 points Two harmonic waves are described by y 1 = A 0 sin(4 x 3 t ) , y 2 = A 0 sin(4 x 3 t φ ) . A 0 = 2 cm, φ = 2. What is the amplitude of the resultant wave? Correct answer: 2 . 16121 cm. Explanation: Since sin θ 1 + sin θ 2 = 2 sin θ 1 + θ 2 2 cos θ 1 θ 2 2 , the resultant wave is y = y 1 + y 2 = 2 A 0 cos φ 2 sin p 4 x 3 t φ 2 P and the amplitude is A = 2 A 0 cos φ 2 = 2 (2 cm) cos 2 2 = 2 . 16121 cm . 005 10.0 points A pulse moves on a string at 1 m/s, traveling to the right. At point A , the string is tightly clamped and cannot move. 1 m/s 1 m A Which of the following shows how the string would look soon after 2 seconds? 1. A 2. A 3. A 4. A
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agnew (ja26385) – oldmidterm 04 – Turner – (58220) 3 5.
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Old Midterm 4 - agnew (ja26385) oldmidterm 04 Turner...

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