Question

A 50-cm-long spring is suspended from the ceiling. A 290 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 26 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?

Answer 1

Solution)

Here, Change in P.E= Change in K.E

Mgh= 0.5 kx^2

0.290*9.8*0.26=0.5*k*0.26^2

So, k= 21.86 N/m (Ans)

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Amplitude= 0.26/2= 0.13 m (Ans)

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Good luck!:)