Question

A body of mass 0.01 kg executes simple harmonic motion (S.H.M.) about x=0 under the influence of a force shown below. The period of the S.H.M. is

A. 1.05 s
B. 0.25 s
C. 0.25 s
D. 0.30 s

Answer 1

Hint:As the time period in simple harmonic motion is defined as the one complete oscillation. So, to solve this question we use the relation of the acceleration in simple harmonic motion which is directly proportional to the displacement.

Formula used:
Force exerted by a body is given as,
\[{f_{\max }} = m{a_{\max }}\]
Where m is the mass of a body and \[{a_{\max }}\] is an acceleration.
The acceleration of a body executes simple harmonic motion (SHM) is given by,
\[{a_{\max }} = - {\omega ^2}x\]
Where \[\omega \] is angular frequency and x is the displacement.
Time period of the S.H.M is given as,
\[T = \dfrac{{2\pi }}{\omega }\]

Complete step by step solution:
Given mass of a body, m=0.01 kg
As we know that \[{f_{\max }} = m{\omega ^2}x\]
From the given graph, at x= 2.0 m and F=8.0 N
By substituting the values, we have
\[8.0 = 0.01 \times {\omega ^2} \times 2.0\]
\[\Rightarrow \omega = 20\,{\rm{ rad/sec}}\]
Hence the period of the S.H.M. can be calculated as,
\[T = \dfrac{{2\pi }}{\omega } \\ \]
\[\therefore T = \dfrac{{2 \times 3.14}}{{20}} = 0.30\,{\rm{ sec}}\]
Therefore, the period of the S.H.M. is 0.30 sec.

Hence option D is the correct answer.

Note: Simple Harmonic Motion (SHM) is defined as a motion in which the restoring force(F) is directly proportional to the displacement(x) of the body from its mean position or equilibrium position. The direction of the restoring force is towards the mean position.
Simple harmonic motion can also be known as an oscillatory motion in which the acceleration of the body at any position is directly proportional to the displacement which is from the mean position.