16 - bhappu (nkb284) Homework 16 florin (58140) 1 This...

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Unformatted text preview: bhappu (nkb284) Homework 16 florin (58140) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points When a body executes simple harmonic mo- tion, its period is 1. independent of its amplitude. correct 2. proportional to its amplitude. 3. proportional to its acceleration. 4. the reciprocal of its speed. 5. inversely proportional to its accelera- tion. Explanation: The period of simple harmonic motion is the same regardless of the amplitude. 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. None of these; the acceleration is con- stant. 2. t = 3 s 3. t = 2 s 4. t = 1 s correct 5. t = 4 s Explanation: This oscillation is described by y ( t ) =- sin t 2 , v ( t ) = dy dt =- 2 cos t 2 a ( t ) = d 2 y dt 2 = 2 2 sin t 2 . The maximum acceleration will occur when sin t 2 = 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 = v t , acceleration is a positive maximum because the velocity is changing from a nega- tive to a positive value. 003 10.0 points Simple harmonic motion can be described us- ing the equation x = x m sin( t + ) . If x = initial position, v = initial velocity, then 1. tan = + v x 2. tan =- v x 3. tan =- x v 4. tan = + x v correct Explanation: x = x m sin( t + ) v = dx dt = x m cos( t + ) When t = 0, x = x m sin bhappu (nkb284) Homework 16 florin (58140) 2 v = x m cos x m sin x m cos = x v tan = x v ....
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16 - bhappu (nkb284) Homework 16 florin (58140) 1 This...

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