solutions 11 - Patel Kinal – Homework 11 – Due 7:00 pm...

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Unformatted text preview: Patel, Kinal – Homework 11 – Due: Oct 12 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A body oscillates with simple harmonic mo- tion along the x-axis. Its displacement varies with time according to the equation x ( t ) = A sin( ω t + φ ) . If A = 5 m, ω = 2 . 684 rad / s, and φ = 1 . 0472 rad, what is the acceleration of the body at t = 3 s? Note: The argument of the sine function is in radians rather than degrees. Correct answer:- 11 . 5211 m / s 2 . Explanation: Let : A = 5 m , ω = 2 . 684 rad / s , φ = 1 . 0472 rad , and t = 3 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 ) =- (2 . 684 rad / s) 2 (5 m) × sin[(2 . 684 rad / s)(3 s) + 1 . 0472 rad] =- 11 . 5211 m / s 2 . keywords: 002 (part 1 of 1) 10 points A body oscillates with simple harmonic mo- tion along the x-axis. Its displacement varies with time according to the equation A = A sin ‡ ω t + π 3 · , where ω = π radians per second, t is in sec- onds, and A = 6 . 2 m. What is the phase of the motion at t = 9 . 6 s? Correct answer: 31 . 2065 rad. Explanation: Let : t = 9 . 6 s and ω = π . x = A sin( ω t + φ ) The phase is the angle in the argument of the sine function, and from the problem state- ment we see it is φ = π t + π 3 = ( π rad / s)(9 . 6 s) + π 3 = 31 . 2065 rad . keywords: 003 (part 1 of 3) 10 points A block of unknown mass is attached to a spring of spring constant 7 . 4 N / m and under- goes simple harmonic motion with an ampli- tude of 8 . 1 cm. When the mass is halfway between its equilibrium position and the end- point, its speed is measured to be 36 . 3 cm / s....
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solutions 11 - Patel Kinal – Homework 11 – Due 7:00 pm...

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