hw14 - karna(pk4534 HW 14 coker(58245 This print-out should...

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karna (pk4534) – HW 14 – coker – (58245) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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 square of the displacement from the equilibrium position. correct 2. the force on a system in stable equilibrium is zero. 3. the force on a system in unstable equilib- rium is zero. 4. the momentum of a system in stable equi- librium is zero. 5. momentum is conserved. 6. the kinetic energy on a system in stable equilibrium is zero. 7. 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. 8. the potential energy of a system near a state of static equilibrium is linearly propor- tional to the displacement from the equilib- rium position. 9. momentum and mechanical energy are both conserved. 10. mechanical energy is conserved. Explanation: For 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 . 002 10.0 points A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. P x t At point P, the mass has 1. negative velocity and positive accelera- tion. 2. positive velocity and negative accelera- tion. correct 3. positive velocity and zero acceleration. 4. negative velocity and negative accelera- tion. 5. positive velocity and positive accelera- tion. 6. negative velocity and zero acceleration. 7. zero velocity and zero acceleration. 8. zero velocity but is accelerating (posi- tively or negatively). Explanation: The velocity is positive because the slope of the curve at P is positive. The acceleration is negative because the curve is concave down at P. 003 10.0 points A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram.
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karna (pk4534) – HW 14 – coker – (58245) 2 1 2 3 4 5 5 5 y (cm) t (s) At what time t in the period shown does the particle achieve its maximum positive ac- celeration? 1. t = 3 s 2. None of these; the acceleration is con- stant. 3. t = 2 s 4. t = 4 s 5. t = 1 s correct Explanation: This oscillation is described by y ( t ) = sin parenleftbigg π t 2 parenrightbigg , v ( t ) = d y dt = π 2 cos parenleftbigg π t 2 parenrightbigg a ( t ) = d 2 y dt 2 = parenleftBig π 2 parenrightBig 2 sin parenleftbigg π t 2 parenrightbigg . The maximum acceleration will occur when sin parenleftbigg π t 2 parenrightbigg = 1, or at t = 1 s . From a non-calculus perspective, the veloc- ity is negative just before t = 1 s since the particle is slowing down. At t = 1 s, the par- ticle is momentarily at rest and v = 0. Just after t = 1 s , the velocity is positive since the particle is speeding up. Remember that a = Δ v Δ t , acceleration is a positive maximum
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