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home / study / questions and answers / science / physics / a 50-cm-long spring is suspended from the ceiling. ... Your question has been answered! Rate it below. Let us know if you got a helpful answer. Question A 50-cm-long spring is suspended from the ceiling. A 310 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 18 cm before coming to rest at its lowest point. It then continues to oscillate vertically. What is the spring constant? What is the amplitude of the oscillation? What is the frequency of the oscillation?

Explanation / Answer

change in GPE = change in SPU

mgh = ½kx²

0.31kg * 9.8m/s² * 0.18m =  ½kx²

0.55 J = ½ * k * (0.18m)² => k = 34 N/m ........spring constant
A = 9 cm ......... amplitude
f = (1/2)(k/m) = (1/2)(34N.m / 0.31kg) = 18.5 Hz ...... frequency

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