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HW #12-solutions - wolz(cmw2833 HW#12 Antoniewicz(57420...

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wolz (cmw2833) – HW #12 – Antoniewicz – (57420) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A 62 . 9 g mass is attached to a horizontal spring with a spring constant of 15 . 4 N / m and released from rest with an amplitude of 24 . 6 cm. What is the velocity of the mass when it is halfway to the equilibrium position if the surface is frictionless? Correct answer: 3 . 3335 m / s. Explanation: Let : m = 62 . 9 g , k = 15 . 4 N / m , and A = 24 . 6 cm . x = A 2 = 0 . 123 m v = radicalbigg k m ( A 2 - x 2 ) = radicalBigg 15 . 4 N / m 0 . 0629 kg [(0 . 246 m) 2 - (0 . 123 m) 2 ] = 3 . 3335 m / s . 002 10.0points The mass of the deuterium molecule D 2 is twice that of the hydrogen molecule H 2 . If the vibrational frequency of H 2 is 1 . 38 × 10 14 Hz, what is the vibrational frequency of D 2 , assuming that the “spring constant” of attracting forces is the same for the two species? Correct answer: 9 . 75807 × 10 13 Hz. Explanation: Let : M D = 2 M H . The angular frequencies depend only on spring constant and mass: ω = radicalbigg k M radicalbigg 1 M ω D = radicalBigg k M D ω H = radicalBigg k M H The spring constants k are the same, so ω D ω H = radicalBigg M H M D = radicalBigg M H 2 M H = radicalbigg 1 2 = 1 2 . The linear frequency is f = ω 2 π ω, so f D f H = ω D ω H = 1 2 f D = f H 2 = 1 . 38 × 10 14 Hz 2 = 9 . 75807 × 10 13 Hz . 003 10.0points At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of bananas into vertical oscillatory motion, which is har- monic with an amplitude 0 . 084 m. The max- imum speed of the bananas is observed to be 0 . 636 m / s. What is the mass of the bananas? The spring of the scale has a force constant 74 . 5 N / m. Correct answer: 1 . 29957 kg. Explanation: Let : A = 0 . 084 m , k = 74 . 5 N / m , and v max = 0 . 636 m / s . A mass on a spring oscillates harmonically according to x ( t ) = A sin( ω t + φ 0 )
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wolz (cmw2833) – HW #12 – Antoniewicz – (57420) 2 so the velocity of the oscillating mass is v ( t ) = d x dt = A ω cos( ω t + φ 0 ) and the maximal speed is | v | max = A ω . Thus the oscillation frequency is ω = | v | max A = radicalbigg k M M = k A 2 | v | 2 max = (74 . 5 N / m)(0 . 084 m) 2 ( v ) 2 = 1 . 29957 kg . 004(part1of4)3.0points An 11 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 ) . Find the angular frequency of the oscillat- ing mass. Correct answer: 95 . 3463 s 1 . Explanation: Let : M = 11 kg and k = 1 × 10 5 N / m . When the mass moves out of equilibrium, it suffers 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 .
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