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Unformatted text preview: Version 081 – Exam 4 – chelikowsky – (59005) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the follow ing is true of the values of its speed and the magnitude of the restoring force? Magnitude of Speed Restoring Force 1. Zero Maximum correct 2. Maximum 1 2 maximum 3. 1 2 maximum 1 2 maximum 4. Maximum Zero 5. Zero Zero Explanation: The maximum displacement occurs at the turning points (where the velocity or speed is zero). The magnitude of restoring force is given by Hooke’s law F = − k x , where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a different perspective, the displace ment from the equilibrium position can be written as y = A sin θ , where θ is the phase of the oscillation. When the object is at its maximum displacement sin θ = 1 θ = π 2 so its speed is v = ω A cos θ = ω A cos π 2 = 0 and the restoring force is F = mA ω 2 sin θ = mA ω 2 π 2 = mA ω 2 , at its maximum value. 002 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. 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 = 2 s 2. t = 1 s correct 3. None of these; the acceleration is con stant. 4. t = 3 s 5. t = 4 s 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 noncalculus perspective, the veloc ity is negative just before t = 1 s since the Version 081 – Exam 4 – chelikowsky – (59005) 2 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 because the velocity is changing from a nega tive to a positive value. 003 (part 1 of 2) 10.0 points A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h . If they are released from rest and roll with out slipping, determine their speeds when they reach the bottom. ( d = disk, h = hoop) 1. v d = radicalbigg 1 2 hg , v h = radicalbigg 1 3 hg 2. v d = radicalbig 2 hg , v h = radicalbig hg 3. v d = radicalbigg 4 3 hg , v h = radicalbig hg correct 4. v d = radicalbig hg , v h = radicalbig 2 hg 5. v d = radicalbigg 2 3 hg , v h = radicalbig 3 hg Explanation: Because they roll without slipping, v = r ω and the total kinetic energy is...
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This note was uploaded on 07/07/2010 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.
 Summer '08
 Staff
 Simple Harmonic Motion

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