Question

A body of mass 0.2kg performs linear S.H.M. It experiences a restoring force of 0.2N when its displacement from the mean position is 4cm and k=5N/m what would be the period of S.H.M ​

Answer 1

Answer:

hi there!!!

Explanation:

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refer to the picture....

Answer 2

Answer:

The time period of the body performing SHM is π sec.

Explanation:

The time period of a particle performing SHM is,

$T=2\pi\sqrt{\frac{m}{k}}$               (1)

Where,

T=time period of the particle

m=mass of the particle

k=spring constant

From the question we have,

m=0.2kg

F=0.2N

x=4cm

k=5N/m

By substituting the value of m and k in equation (1) we get;

$T=2\pi\sqrt{\frac{0.2}{5}}=2\pi\sqrt{\frac{1}{4}} =\frac{2\pi}{2}=\pi sec$

Hence, the time period of the body performing SHM is π sec.