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Unformatted text preview: zhou (cz3574) HW13 mackie (10611) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A simple harmonic oscillator takes 11 . 5 s to undergo five complete vibrations. Find the period of its motion. Correct answer: 2 . 3 s. Explanation: Let : t = 11 . 5 s and n = 5 . The period is, by definition, how long it takes the oscillator to complete one full cycle; i.e. , to return to where it started. Since we are told how long it takes to undergo five complete vibrations, the period is T = t n = 11 . 5 s 5 = 2 . 3 s . 002 (part 2 of 3) 10.0 points Calculate the frequency. Correct answer: 0 . 434783 Hz. Explanation: The frequency is how many vibrations the oscillator undergoes per unit time, or the re ciprocal of the period: f = 1 T = 1 2 . 3 s = . 434783 Hz 003 (part 3 of 3) 10.0 points Find the angular frequency. Correct answer: 2 . 73182 rad / s. Explanation: = 2 f = 2 (0 . 434783 Hz) = 2 . 73182 rad / s . 004 10.0 points Military specifications often call for electronic devices to be able to withstand accelerations of 10 g . To make sure that their products meet this specification, manufacturers test them using a shaking table that can vibrate a device at various specified frequencies and amplitudes. If a device is given a vibration of amplitude 5 . 4 cm, what should be its frequency in order to test for compliance with the 10 g military specification? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 6 . 78356 Hz. Explanation: Let : a max = 10 g = 98 . 1 m / s 2 and A = 5 . 4 cm = 0 . 054 m . The angular frequency is = 2 f and the maximum accleration of the oscillator is a max = A 2 = 4 2 A f 2 f = 1 2 radicalbigg a max A = 1 2 radicalbigg 98 . 1 m / s 2 . 054 m = 6 . 78356 Hz ....
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 Spring '10
 Hiyishi

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