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Unformatted text preview: Version 081 – Exam 4 – chelikowsky – (59005) 1 This print-out should have 22 questions. Multiple-choice 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 non-calculus 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
- Simple Harmonic Motion