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Unformatted text preview: yang (ey942) 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( t + ) . If A = 3 m, = 3 . 691 rad / s, and = 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: 4 . 21174 m / s 2 . Explanation: Let : A = 3 m , = 3 . 691 rad / s , = 1 . 0472 rad , and t = 4 s . x = A sin( t + ) v = dx dt = A cos( t + ) a = dv dt = 2 A sin( t + ) = 2 A sin( t + ) = (3 . 691 rad / s) 2 (3 m) sin[(3 . 691 rad / s)(4 s) + 1 . 0472 rad] = 4 . 21174 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 . 36 m. At a point 2 . 2176 m away from the equilibrium, the mass has speed 3 . 47 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 . Correct answer: 4 . 5707 s. Explanation: Let A = 3 . 36 m , x = 2 . 2176 m , and v = 3 . 47 m / s . The simplest solution uses the equation for simple harmonic motion x ( t ) = A sin( t ) , and its time derivative v ( t ) = dx dt = A cos( t ) . For any angle (such as the phase angle t ) sin 2 ( t ) + cos 2 ( t ) = 1 , so parenleftBig x A parenrightBig 2 + parenleftBig v A parenrightBig 2 = 1 , and parenleftBig v parenrightBig 2 = A 2 x 2 . Consequently, = v A 2 x 2 = 3 . 47 m / s radicalbig (3 . 36 m) 2 (2 . 2176 m) 2 = 1 . 37466 s 1 and T = 2 = 2 1 . 37466 s 1 = 4 . 5707 s . 003 (part 1 of 2) 10.0 points A 1 . 93 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4 . 52 N / m. The object is displaced 2 . 35 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 . 88 s after it is released? Correct answer: 9 . 99456 N. Explanation: yang (ey942) HW11 TSOI (58160) 2 Let : A = 2 . 35 m , k = 4 . 52 N / m , m = 1 . 93 kg , and t = 3 . 88 s . The block exhibits simple harmonic motion, so the displacement is x = A cos( t ) where the angular frequency is = radicalbigg k m . Thus the force is F = k x = k A cos parenleftBigg radicalbigg k m t parenrightBigg = (4 . 52 N / m) (2 . 35 m) cos bracketleftBigg radicalBigg 4 . 52 N / m 1 . 93 kg (3 . 88 s) bracketrightBigg = 9 . 99456 N a force of 9 . 99456 N directed to the left....
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This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas at Austin.
 Fall '10
 TSIO

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