A body of mass 4kg hangs from a spring and oscillates with a period 0.5 seconds. On removal of the body the spring shortens by how much?

Question

Saurabh Kumar · Answer

At equilibrium, mg = kx eqv X eqv = mg/k Restoring force is developed due to application of extra force during extension.

Rahul shah · Answer

Hello, We know that, T=2π(√m/k) Putting the value of T, m, after we get K=64x9.9~64x10 Then use formula.kx=mg. X=6.2cm

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A body of mass 4kg hangs from a spring and oscillates with a period 0.5 seconds. On removal of the body the spring shortens by how much?

A body of mass 4kg hangs from a spring and oscillates with a period 0.5 seconds. On removal of the body the spring shortens by how much?

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A body of mass 4kg hangs from a spring and oscillates with a period 0.5 seconds. On removal of the body the spring shortens by how much?

A body of mass 4kg hangs from a spring and oscillates with a period 0.5 seconds. On removal of the body the spring shortens by how much?

2 Answers

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