Question

A body of mass 2 g performs linear SHM .If the restoring force acting on it is 3N when it is 0.06m from the mean position ,

the period of the SHM
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Answer 1

Given : -

Mass of body = 2g = 0.002kg

Restoring force = 3N

Distance from mean position = 0.06m

To Find : -

Time period of the SHM.

Solution : -

First of all we need to find constant k.

We know that restoring force F is directly proportional to the distance of body from mean position. [F ∝ x]

By putting constant,

$\longrightarrow \sf F = kx$

By substituting given data,

$\longrightarrow \sf 3 = k \times 0.06 \\\\\longrightarrow k = 3/0.06\\\\ \longrightarrow k = 300/6 \\\\\longrightarrow k = 50 N/m$

Time period of SHM is given by

$\longrightarrow \sf T = 2\pi \sqrt {\dfrac{m}{k}} \\\\ \longrightarrow T = 2\times 3.14\times \sqrt{\dfrac{0.002}{50} } \\\\\longrightarrow T = 6.28 \times \sqrt{0.00004} \\\\\longrightarrow T = 6.28 \times 0.0063 \\\\\longrightarrow \bf T = 0.04s$