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Unformatted text preview: This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 9 . 1 kg mass slides on a frictionless surface and is attached to two springs with spring constants 37 N / m and 79 N / m as shown in the figure. 37 N / m 79 N / m 9 . 1 kg Find the frequency of oscillation. Correct answer: 0 . 568236 Hz. Explanation: Let : k 1 = 37 N / m , k 2 = 79 N / m , and m = 9 . 1 kg . By Hooke’s law, F =- k x = ma = m d 2 x dt 2 and d 2 x dt 2 + k m x = 0 , (1) whose integral form has a sine function x ( t ) = A sin( ω t + δ ) , where ω = radicalbigg k m , the square root of the coeffi- cient of x in Eq. 1. Call the displacement of the mass x , and choose the positive direction to be to the right. Then, the forces from the springs on the mass m are to the left: F 1 =- k 1 x and F 2 =- k 2 x , so that force equilibrium is F =- k parallel x F 1 + F 2 =- k parallel x- k 1 x- k 2 x =- k parallel x k parallel = k 1 + k 2 . ω = radicalbigg k m , so ω parallel = radicalbigg k parallel m = radicalbigg k 1 + k 2 m and the frequency of oscillation is f = ω parallel 2 π = radicalbigg k 1 + k 2 m 2 π = radicalbigg 37 N / m + 79 N / m 9 . 1 kg 2 π = . 568236 Hz . 002 (part 2 of 2) 10.0 points Now the two spring are coupled together as shown in the figure. 37 N / m 79 N / m 9 . 1 kg Find the period of oscillation....
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