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Unformatted text preview: johnson (mjj622) – HW 14 – Coker – (56625) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This assignment covers Chapter 15 on the Simple Harmonic Oscillator. 001 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. zero velocity but is accelerating (posi tively or negatively). 2. positive velocity and zero acceleration. 3. positive velocity and negative accelera tion. correct 4. negative velocity and zero acceleration. 5. negative velocity and positive accelera tion. 6. negative velocity and negative accelera tion. 7. positive velocity and positive accelera tion. 8. zero velocity and zero acceleration. 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. 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. None of these; the acceleration is con stant. 3. t = 1 s correct 4. t = 4 s 5. t = 3 s 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. 003 (part 1 of 4) 10.0 points johnson (mjj622) – HW 14 – Coker – (56625) 2 A 0 . 2 kg block attached to a spring of force constant 19 . 6 N / m oscillates with an ampli tude of 15 cm. Find the maximum speed of the block. Correct answer: 1 . 48492 m / s. Explanation: Let : A = 15 cm = 0 . 15 m , k = 19 . 6 N / m , and m = 0 . 2 kg . v max = Aω = A radicalbigg k m = (0 . 15 m) radicalBigg 19 . 6 N / m . 2 kg = 1 . 48492 m / s . 004 (part 2 of 4) 10.0 points Find the speed of the block when it is 7 . 5 cm from the equilibrium position. Correct answer: 1 . 28598 m / s. Explanation: Let : x = 7 . 5 cm = A 2 ....
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This note was uploaded on 02/10/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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