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Unformatted text preview: salazar (ggs332) – HW11 – TSOI – (58160) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A body oscillates with simple harmonic mo tion along the xaxis. Its displacement varies with time according to the equation x ( t ) = A sin( ω t + φ ) . If A = 3 m, ω = 3 . 572 rad / s, and φ = 1 . 0472 rad, what is the acceleration of the body at t = 4 s? Note: The argument of the sine function is in radians rather than degrees. Correct answer: − 13 . 9404 m / s 2 . Explanation: Let : A = 3 m , ω = 3 . 572 rad / s , φ = 1 . 0472 rad , and t = 4 s . x = A sin( ω t + φ ) v = dx dt = ω A cos( ω t + φ ) a = dv dt = − ω 2 A sin( ω t + φ ) = − ω 2 A sin( ω t + φ ) = − (3 . 572 rad / s) 2 (3 m) × sin[(3 . 572 rad / s)(4 s) + 1 . 0472 rad] = − 13 . 9404 m / s 2 . 002 10.0 points A mass attached to a spring executes sim ple harmonic motion in a horizontal plane with an amplitude of 3 . 64 m. At a point 1 . 3468 m away from the equilibrium, the mass has speed 0 . 886 m / s. What is the period of oscillation of the mass? Consider equations for x ( t ) and v ( t ) and use sin 2 +cos 2 = 1 to calculate ω . Correct answer: 23 . 9816 s. Explanation: Let A = 3 . 64 m , x = 1 . 3468 m , and v = 0 . 886 m / s . The simplest solution uses the equation for simple harmonic motion x ( t ) = A sin( ω t ) , and its time derivative v ( t ) = dx dt = Aω cos( ω t ) . For any angle (such as the phase angle ω t ) sin 2 ( ω t ) + cos 2 ( ω t ) = 1 , so parenleftBig x A parenrightBig 2 + parenleftBig v A ω parenrightBig 2 = 1 , and parenleftBig v ω parenrightBig 2 = A 2 − x 2 . Consequently, ω = v √ A 2 − x 2 = . 886 m / s radicalbig (3 . 64 m) 2 − (1 . 3468 m) 2 = 0 . 262 s − 1 and T = 2 π ω = 2 π . 262 s − 1 = 23 . 9816 s . 003 (part 1 of 2) 10.0 points A 1 . 91 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4 . 59 N / m. The object is displaced 4 . 4 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3 . 67 s after it is released? Correct answer: 16 . 7374 N. Explanation: salazar (ggs332) – HW11 – TSOI – (58160) 2 Let : A = 4 . 4 m , k = 4 . 59 N / m , m = 1 . 91 kg , and t = 3 . 67 s . The block exhibits simple harmonic motion, so the displacement is x = A cos( ω t ) where the angular frequency is ω = radicalbigg k m . Thus the force is F = − k x = − k A cos parenleftBigg radicalbigg k m t parenrightBigg = − (4 . 59 N / m) (4 . 4 m) cos bracketleftBigg radicalBigg 4 . 59 N / m 1 . 91 kg (3 . 67 s) bracketrightBigg = − 16 . 7374 N a force of 16 . 7374 N directed to the left....
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This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics, Simple Harmonic Motion

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