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Unformatted text preview: obennoskey (blo242) – hw11 – Demkov – (59910) 1 This printout should have 44 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 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 = 1 s correct 2. t = 2 s 3. t = 3 s 4. t = 4 s 5. None of these; the acceleration is con stant. Explanation: This oscillation is described by y ( t ) = sin parenleftbigg π t 2 parenrightbigg , v ( t ) = dy 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 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. 002 10.0 points The equation of motion of a simple harmonic oscillator is d 2 x dt 2 = 9 x, where x is displacement and t is time. What is the period of oscillation? 1. T = 3 2 π 2. T = 9 2 π 3. T = 2 π 3 correct 4. T = 2 π 9 5. T = 6 π Explanation: d 2 x dt 2 = ω 2 x, where ω is the angular frequency, so the pe riod of oscillation is T = 2 π ω = 2 π √ 9 = 2 π 3 . 003 10.0 points A block rests on a flat plate that executes vertical simple harmonic motion with a period of 1 . 82 s. What is the maximum amplitude of the motion for which the block does not separate from the plate? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 82226 m. Explanation: Let : g = 9 . 8 m / s 2 , and T = 1 . 82 s . obennoskey (blo242) – hw11 – Demkov – (59910) 2 At each instant, there are two forces acting on the block: the force of gravity F g = mg and the normal force N from the plate. The block separates from the plate when N = 0. By Newton 2 nd law summationdisplay F = mg + N = ma, where up is positive. We see that N = m ( g + a ) , so for N to vanish, we must have a max = g , (1) For simple harmonic motion x = A cos( ω t ) a = d 2 x dt 2 = Aω 2 cos( ω t ) , and at the highest and lowest points, the ac celeration is at a maximum. The acceleration of the block is positive (up) at the lowest point, and negative (down) at the highest point. We want the negative situation, so a max = A max ω 2 ....
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This note was uploaded on 12/04/2008 for the course PHY PHY301 taught by Professor Demovk during the Spring '08 term at University of Texas at Dallas, Richardson.
 Spring '08
 DEMOVK

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