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HW11 Solution - salazar(ggs332 HW11 TSOI(58160 This...

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salazar (ggs332) – HW11 – TSOI – (58160) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 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 = 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 ) = dt = cos( ω t ) . ±or any angle (such as the phase angle ω t ) sin 2 ( ω t ) + cos 2 ( ω t ) = 1 , so p x A P 2 + p v A ω P 2 = 1 , and p v ω P 2 = A 2 x 2 . Consequently, ω = v A 2 x 2 = 0 . 886 m / s r (3 . 64 m) 2 (1 . 3468 m) 2 = 0 . 262 s 1 and T = 2 π ω = 2 π 0 . 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:
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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 = r k m . Thus the force is F = k x = k A cos p r k m t P = (4 . 59 N / m) (4 . 4 m) cos b R 4 . 59 N / m 1 . 91 kg (3 . 67 s) B = 16 . 7374 N a force of 16 . 7374 N directed to the left. 004 (part 2 of 2) 10.0 points How many times does the object oscillate in 3 . 67 s? Correct answer: 0 . 905474 turn. Explanation: The angular frequency is w = r k m and the period of oscillation is T = 2 π ω = 2 π r m k , so the number of oscillations made in 3 . 67 s is N = Δ t T = Δ t 2 π r k m = 3 . 67 s 2 π R 4 . 59 N / m 1 . 91 kg = 0 . 905474 turn . 005 10.0 points A uniform plank of mass m is pivoted at one end. A spring of force constant k is attached to the center of the plank, as shown in the Fgure. The height of the pivot has been adjusted so that the plank will be in equilibrium when it is horizontally oriented.
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HW11 Solution - salazar(ggs332 HW11 TSOI(58160 This...

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