Question #226088

A 5.0-{\rm kg} block is suspended from the ceiling by a strong spring and released to perform simple harmonic motion with a period of 0.65 {\rm \;s} . The block is brought to rest, and the length of the spring with the block attached is measured. By how much is this length reduced when the block is removed?


Expert's answer

The period of a spring pendulum is


T=2πmk, k=4π2mT2=467.2 N/m.T=2\pi\sqrt{\frac mk},\\\space\\ k=\frac{4\pi^2m}{T^2}=467.2\text{ N/m}.

According to equilibrium of forces:


mg=kx, x=mgk=0.1049 m.mg=kx,\\\space\\x=\frac{mg}{k}=0.1049\text{ m}.


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