1169061395Xuj

1169061395Xuj - Alvarado, Patrick Homework 12 Due: Dec 14...

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Alvarado, Patrick – Homework 12 – Due: Dec 14 2006, 10:00 am – Inst: Andrei Sirenko 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 2) 10 points A block on a horizontal Frictionless plane is attached to a spring, as shown below. The block oscillates along the x -axis with simple harmonic motion oF amplitude A . m k 0 - A + A v x 0 Which oF the Following statements about the block is correct? 1. At x = 0, its velocity is zero. 2. At x = A , its acceleration is zero. 3. At x = A , its velocity is at a maximum. 4. At x = 0, its acceleration is at a maxi- mum. 5. At x = A , its displacement is at a maxi- mum. correct Explanation: The block oscillates From maximum dis- placements at x = A and x = - A . At those points the velocity is momentarily zero. 002 (part 2 oF 2) 10 points Which oF the Following statements about en- ergy is correct? 1. The kinetic energy oF the block is at a minimum at x = 0. 2. The potential energy oF the spring is at a minimum at x = 0. correct 3. The kinetic energy oF the block is at a maximum at x = A . 4. The potential energy oF the spring is at a minimum at x = A . 5. The kinetic energy oF the block is always equal to the potential energy oF the spring. Explanation: ±rom conservation oF energy, v = 0 at x = ± A so the kinetic energy is zero and the spring potential energy is at its maximum. At x = 0, the spring potential energy is 0 and the kinetic energy is at its maximum. keywords: 003 (part 1 oF 1) 10 points A particle oscillates harmonically x = A sin( ωt + φ 0 ) , with amplitude A = 21 m, angular Frequency ω = π s - 1 , and initial phase φ 0 = π 3 radians. Every now and then, the particle’s kinetic energy and potential 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. 0.1294 s 3. 0.5884 s 4. 0.2238 s 5. 0.7615 s 6. 0.9967 s 7. 0.5167 s 8. 0.8623 s 9. 0.6547 s 10. 0.3467 s Explanation: Basic Concepts: x = A cos( ω t + φ )
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Alvarado, Patrick – Homework 12 – Due: Dec 14 2006, 10:00 am – Inst: Andrei Sirenko 2 K = 1 2 mv 2 sin 2 ( ω t + δ ) U = 1 2 mx 2 = cos 2 ( ω t + δ ) p sin( π t + π/ 3) = p cos( π t + π/ 3) | sin( π t + π/ 3) | = | cos( π t + π/ 3) | . We note that | sin( x ) | = | cos( x ) | when x = (2 n + 1) π 4 , where n is an integer. Hence (2 n + 1) π 4 = π t + π 3 (2 n + 1) 4 = t + 1 3 , so t = (2 n + 1) 4 - 1 3 = 0 . 416667 s . Note: When n = 0, we obtain a negative value for time. The Frst positive answer is when n = 1. keywords: 004 (part 1 of 1) 10 points The displacement in simple harmonic motion is a maximum when 1. the potential energy function has its max- imum value per cycle. correct 2. kinetic energy is maximum. 3.
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1169061395Xuj - Alvarado, Patrick Homework 12 Due: Dec 14...

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