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A spring stretches by 0.0226 m when a 3.74-kg object is suspended from its end. How...

A spring stretches by 0.0226 m when a 3.74-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 5.53 Hz?

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Answer #1

Gravitational acceleration = g = 9.81 m/s2

Force constant of the spring = k

Mass of the first object = m1 = 3.74 kg

Amount the spring stretches due to the first object = X1 = 0.0226 m

m1g = kX1

(3.74)(9.81) = k(0.0226)

k = 1623.425 N/m

Mass to be attached to the spring for a frequency of vibration to be 5.53 Hz = m2

Frequency of vibration = f = 5.53 Hz

m2 = 1.344 kg

Mass of the object to be attached to the spring so that its frequency of vibration is 5.53 Hz = 1.344 kg

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