# HW 14 - johnson(mjj622 – HW 14 – Coker –(56625 1 This...

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Unformatted text preview: johnson (mjj622) – HW 14 – Coker – (56625) 1 This print-out should have 22 questions. Multiple-choice 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 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 = 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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HW 14 - johnson(mjj622 – HW 14 – Coker –(56625 1 This...

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