Question

The max. speed of a particle in S.H.M. is found to be $62.8cm/s$. If the amplitude is 20cm its period of oscillation is
(A) 1 sec
(B) 2 sec
(C) 3 sec
(D) 4 sec

Answer 1

Hint
From the relation between the angular speed of a particle and the maximum velocity in an S.H.M. ${v_{\max }} = A\omega $ , we can find the angular speed. Then, from the angular speed we can find the time period of oscillation.

Formula Used: In this solution, we will be using the following formula,
$\Rightarrow {v_{\max }} = A\omega $
where ${v_{\max }}$ is the maximum speed of the particle in S.H.M.
$A$ is the amplitude of the oscillation of the S.H.M. and $\omega $ is the angular velocity.
where angular velocity $\omega = \dfrac{{2\pi }}{T}$, $T$ is the time period of oscillation.

Complete step by step answer
For an S.H.M. The maximum speed is given by the product of the amplitude and the angular speed of the wave.
So ${v_{\max }} = A\omega $.
Now in the question, we are provided with the maximum speed and the amplitude of the wave. So from there, we can calculate the angular speed of the wave.
Therefore we can write from the previous equation,
$\Rightarrow \omega = \dfrac{{{v_{\max }}}}{A}$
Substituting the values ${v_{\max }} = 62.8cm/s$ and $A = 20cm$in the equation we get,
$\omega = \dfrac{{62.8}}{{20}}rad/s$
Calculating this we get,
$\Rightarrow \omega = 3.14rad/s$
The value $3.14$ is approximately equal to the value of $\pi $.
So the value of angular speed as,
$\Rightarrow \omega = \pi rad/s$
Now, the angular speed and the time period are related by the equation,
$\Rightarrow \omega = \dfrac{{2\pi }}{T}$
We can write this as,
$\Rightarrow T = \dfrac{{2\pi }}{\omega }$
Now substituting the value of angular speed in this equation we get,
$\Rightarrow T = \dfrac{{2\pi }}{\pi }s$
Therefore, we get,
$\Rightarrow T = 2s$
So the time period of oscillation is 2 sec.
Hence the correct option is B.

Note
The simple harmonic motion is a special type of periodic motion. Here the displacement of the object is directly proportional to its restoring force and it acts towards the equilibrium position of the object. In an S.H.M. at the equilibrium position, the speed is zero and it is the maximum at the extremes.