This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Niu (qn269) – HW10 – Tsoi – (58020) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find 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. Find 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 radicalbigg 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 = radicalbigg 2 E k = radicalBigg 2 (4 . 2 J) 41 . 871 N / m = . 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: For simple harmonic motion, the restoring force obeys Hooke’s law ( F ∝ x − x , the displacement from the equilibrium position), so the potential energy U ∝ ( x − x ) 2 . 003 10.0 points A particle oscillates harmonically x = A sin( ωt + φ ) , 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 first time after t = 0? 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 = π 3 . x = A cos( ω t + φ ) v = − Aπ sin( ω t + φ ) and ω = radicalbigg k m . K = U 1 2 mv 2 = 1 2 k x 2 A 2 ω 2 sin 2 ( ω t + φ ) = k m A 2 cos 2 ( ω t + φ ) ω radicalBig sin 2 ( ω t + φ ) = radicalbigg k m radicalBig cos 2 ( ω t + φ ) vextendsingle vextendsingle vextendsingle sin parenleftBig π t + π 3 parenrightBigvextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle cos parenleftBig π t + π 3 parenrightBigvextendsingle vextendsingle vextendsingle ....
View
Full
Document
This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas.
 Spring '09
 Staff
 Physics, Mass

Click to edit the document details