hw12 - strong (ams4684) HW12 ditmire (58216) This print-out...

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strong (ams4684) – HW12 – ditmire – (58216) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 4) 10.0 points A3 . 9 kg mass is suspended on a 1 × 10 5 N / m spring. The mass oscillates up and down From the equilibrium position y eq = 0 according to y ( t )= A sin( ωt + φ 0 ) . ±ind the angular Frequency oF the oscillat- ing mass. Correct answer: 160 . 128 s - 1 . Explanation: Let : M =3 . 9 kg and k =1 × 10 5 N / m . When the mass moves out oF equilibrium, it su²ers a net restoring Force F net y = F spring - Mg = - k ( y - y eq - ky , and accelerates back towards the equilibrium position at the rate a y = F net y M = - k M y. ThereFore, the mass oscillates harmonically with angular Frequency ω = ± - a y y = ± k M = ² 1 × 10 5 N / m 3 . 9 kg = 160 . 128 s - 1 . 002 (part 2 oF 4) 10.0 points At time t 0 = 0 the mass is at 17 . 5 cm and moving upward at velocity +44 . 9m / s. ±ind the amplitude oF the oscillating mass. Correct answer: 33 . 0529 cm. Explanation: Let : y 0 = 17 . 5 cm and v 0 = 44 . / s . The mass oscillates according to the SHM equation y ( t A sin( + φ 0 ) , so its velocity is v y ( t dy dt = A ω cos( + φ 0 ) . At time t 0 = 0, we have y 0 y ( t = 0) = A sin φ 0 and v 0 = v y ( t = 0) = cos φ 0 , so A sin φ 0 = y 0 and A cos φ 0 = v 0 ω . (1) Consequently, A 2 =( A sin φ 0 ) 2 +( A cos φ 0 ) 2 y 0 ) 2 + ³ v 0 ω ´ 2 A = ± ( y 0 ) 2 + ³ v 0 ω ´ 2 = ² (17 . 5 cm) 2 + µ 44 . / s 160 . 128 s - 1 2 = 33 . 0529 cm . 003 (part 3 oF 4) 10.0 points ±ind the initial phase φ 0 in the range 0 φ 0 < 360 . Correct answer: 31 . 9686 . Explanation: According to eq. (1), tan φ 0 = A sin φ 0 A cos φ 0 = ω y 0 v 0 =0 . 624108 . y 0 > 0 and v 0 > 0 imply sin φ 0 > 0 , cos φ 0 > 0 , and 0 < φ 0 < 90 . Consequently, the initial phase is φ 0 = arctan(0 . 624108) = 31 . 9686 .
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strong (ams4684) – HW12 – ditmire – (58216) 2 004 (part 4 of 4) 10.0 points Calculate the position of the oscillating mass at the time t =0 . 0371 s . Correct answer: 7 . 06875 cm. Explanation: Let : t . 0371 s . y = A sin( ωt + φ 0 ) = (33 . 0529 cm) × sin[(160 . 128 s - 1 )(0 . 0371 s) + 31 . 9686 ] = 7 . 06875 cm . 005 (part 1 of 2) 10.0 points A6 . 3 kg mass slides on a frictionless surface and is attached to two springs with spring constants 23 N / m and 65 N / m as shown in the Fgure. 23 N / m 65 N / m 6 . 3 kg ±ind the frequency of oscillation. Correct answer: 0 . 594828 Hz. Explanation: Let : k 1 = 23 N / m , k 2 = 65 N / m , and m =6 . 3 kg . By Hooke’s law, F = - kx = ma = m d 2 x dt 2 and d 2 x dt 2 + k m x , (1) whose integral form has a sine function x ( t )= A sin( ω t + δ ) , where ω = ± k m , the square root of the coe²- 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 = ± k m , so ω parallel = ± k parallel m = ± k 1 + k 2 m and the frequency of oscillation is f = ω parallel 2 π = ± k 1 + k 2 m 2 π = ± 23 N / m + 65 N / m 6 . 3 kg 2 π = 0 . 594828 Hz .
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hw12 - strong (ams4684) HW12 ditmire (58216) This print-out...

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