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Unformatted text preview: lee (cyl327) – HW11 – gentle – (56245) 1 This printout should have 8 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A simple harmonic oscillator has amplitude . 94 m and period 3 sec. What is the maximum acceleration? 1. . 656244 m / s 2 2. 2 . 06165 m / s 2 3. 12 . 3699 m / s 2 4. . 104444 m / s 2 5. 4 . 1233 m / s 2 correct 6. . 313333 m / s 2 Explanation: Let : A = 0 . 94 m and T = 3 sec . For a simple harmonic oscillator, the dis placement is x = A cos parenleftbigg 2 π T t + φ parenrightbigg , so the acceleration is a = d 2 x dt 2 = A parenleftbigg 2 π T parenrightbigg 2 cos parenleftbigg 2 π T t + φ parenrightbigg . Since 1 < cos α < 1, the maximum acceler ation is A max = 4 π 2 A T 2 = 4 π 2 (0 . 94 m) (3 sec) 2 = 4 . 1233 m / s 2 . 002 10.0 points Simple harmonic motion can be described us ing the equation x = x m sin( ωt + φ ) . If x = initial position, v = initial velocity, then 1. tan φ = ω x v 2. tan φ = + ω x v correct 3. tan φ = v ω x 4. tan φ = + v ω x Explanation: x = x m sin( ω t + φ ) v = dx dt = x m ω cos( ω t + φ ) When t = 0, x = x m sin φ v = x m ω cos φ x m sin φ x m ω cos φ = x v tan φ = ω x v . 003 10.0 points A 10 kg person steps into a car of mass 2606 kg, causing it to sink 0 . 28 cm on its springs. Assuming no damping, with what fre quency will the car and passenger vibrate on the springs? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 0 . 582447 Hz. Explanation: Let : m = 10 kg , M = 2606 kg , g = 9 . 81 m / s 2 , and Δ x = 0 . 28 cm = 0 . 0028 m . lee (cyl327) – HW11 – gentle – (56245) 2 The spring force is F = k Δ x k = F Δ x = mg Δ x , where m is the person’s mass....
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.
 Fall '10
 MARKERT
 mechanics, Simple Harmonic Motion

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