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Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 2.7 cm. You observe that it takes 2.8 s to make one round trip. You construct another vertical oscillator with a mass 3 times as heavy and a spring 4 times as stiff. You take it to a planet where 2.4 N/kg. You start it oscillating with amplitude 4.9 cm. How long does it take for the mass to make one round trip

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Answer:

The period of an oscillator in simple does not depend on gravity as one can see from measurement of a horizontal spring.

Let w be angular frequencu (omega)

w = (k / m)^1/2 for an object in SHM

w2 = (4 k / 3 m)^1/2  = 1.15 (k / m)^1/2 = 1.15 w

The angular frequency for the new oscillator is 1.15 that of the original

w = 2 pi * f = 2 pi / T    

the period T = 2 pi / w = 2.8 s (given)

T2 / T = w / w2 = 1 / 1.15 = .87

T2 = .87 * 2.8 = 2.43 s

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