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A copper rod (length = 2.45 m, radius = 3.17 x 10-3 m, Young's modulus 1.1 x 1011 N/m2) hangs down from the ceiling. A 9.33-kg object is attached to the lower end of the rod. The rod acts as a "spring," and the object oscillates vertically with a small amplitude. Ignoring the rod's mass, find the frequency f of the simple harmonic motion.
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Youngs modlulus Y of a material is given Y = FL/ADL

A = area

DL = change of length

L = original length

F = force

so DL = FL/AY

DL = 9.33*9.81* 2.45/(3.14*0.00317*0.00317*1.1*10^11)

DL = 0.0000646m

so now use frequency f = (1/2L)(sqrt(K/m)

where K = Fx = 9.33* 9.8* 0.0000646 = 0.0059

so f = 1/(2*2.45)*sqrt(0.0059/9.33)

f = 0.0049 Hz
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