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hw10solutions

# hw10solutions - Niu(qn269 HW10 Tsoi(58020 This print-out...

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Niu (qn269) – HW10 – Tsoi – (58020) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A 0 . 462 kg mass is attached to a spring and undergoes simple harmonic motion with a pe- riod oF 0 . 66 s. The total energy oF the system is 4 . 2 J. ±ind the amplitude oF the motion. Correct answer: 0 . 447902 m. Explanation: Let : m = 0 . 462 kg , T = 0 . 66 s , and E = 4 . 2 J . The period is T = 2 π r m k k = 4 π 2 m T 2 = 4 π 2 (0 . 462 kg) (0 . 66 s) 2 = 41 . 871 N / m , and the energy is E = 1 2 k A 2 A = r 2 E k = R 2 (4 . 2 J) 41 . 871 N / m = 0 . 447902 m . 002 10.0 points IF almost any system in stable equilibrium is slightly disturbed, it will then exhibit simple harmonic motion because 1. the potential energy oF a system near a state oF static equilibrium is proportional to the cube oF the displacement From the equi- librium position. 2. momentum and mechanical energy are both conserved. 3. mechanical energy is conserved. 4. the Force on a system in stable equilibrium is zero. 5. the momentum oF a system in stable equi- librium is zero. 6. the potential energy oF a system near a state oF static equilibrium is linearly propor- tional to the displacement From the equilib- rium position. 7. the potential energy oF a system near a state oF static equilibrium is proportional to the square oF the displacement From the equilibrium position. correct 8. momentum is conserved. 9. the kinetic energy on a system in stable equilibrium is zero. 10. the Force on a system in unstable equilib- rium is zero. Explanation: ±or simple harmonic motion, the restoring Force obeys Hooke’s law ( F x x 0 , the displacement From the equilibrium position), so the potential energy U ( x x 0 ) 2 . 003 10.0 points A particle oscillates harmonically x = A sin( ωt + φ 0 ) , with amplitude 13 m, angular Frequency π s 1 , and initial phase 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?

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Niu (qn269) – HW10 – Tsoi – (58020) 2 1. 0.4167 s correct 2. 0.9967 s 3. 0.7615 s 4. 0.6547 s 5. 0.5884 s 6. 0.8623 s 7. 0.5267 s 8. 0.2238 s 9. 0.3467 s 10. 0.1294 s Explanation: Let : A = 13 m , ω = π s 1 , and φ 0 = π 3 . x = A cos( ω t + φ 0 ) v = A π 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 004 10.0 points If almost any system in stable equilibrium is slightly disturbed, it will then exhibit simple harmonic motion because 1. the force on a system in stable equilibrium is zero. 2. the change in potential energy is propor- tional to the square of the displacement from the equilibrium position. correct 3. momentum is conserved.
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hw10solutions - Niu(qn269 HW10 Tsoi(58020 This print-out...

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