Question

The time period of oscillation of a particle that executes SHM is 1.2s. The time starting from the mean position at which its velocity will be half of its velocity at mean position is?
A. 0.1s
B. 0.2s
C. 0.4s
D. 0.6s

Answer 1

Hint: You could recall the expression for velocity in an SHM and hence find the velocity at mean position. After that you could go for the point at which the velocity is one half that at the mean position. There you get the displacement from the mean position. Substitute this in expression of displacement and also substitute for time period to get the answer.

Formula used: Displacement,
$x=A\cos \omega t$
Velocity,
$u=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$
Angular velocity,
$\omega =\dfrac{2\pi }{T}$

Complete step by step answer:
In the question, we are given the time period of oscillation of particle executing SHM as 1.2s. We are supposed to find the time taken to cover the distance from the mean position to the point at which the velocity is half of that value at mean position.
We know that the velocity is given by,
$u=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$ …………………………….. (1)
At the mean position the velocity of the particle would be,
${{u}_{\max }}=\omega A$ …………………………………. (2)
Now, the velocity at the point where the velocity is half that at the mean position is given by,
$u=\dfrac{{{u}_{\max }}}{2}=\dfrac{\omega A}{2}=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$
$\Rightarrow \dfrac{{{A}^{2}}}{4}={{A}^{2}}-{{x}^{2}}$
$\Rightarrow x=\dfrac{\sqrt{3}}{2}A$ ……………………………………. (3)
Now, we know that the displacement of a particle at a particular point is given by,
$x=A\cos \omega t$
Substituting (3) we get,
$\dfrac{\sqrt{3}}{2}A=A\cos \omega t$
$\Rightarrow \omega t={{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}$
$\Rightarrow t=\dfrac{\pi }{6\omega }$
But we know,
$\omega =\dfrac{2\pi }{T}=\dfrac{2\pi }{1.2}$
$\Rightarrow t=\dfrac{1.2\pi }{12\pi }$
$\therefore t=0.1\sec $

So, the correct answer is “Option A”.

Note: Points to be remembered about SHM is that the mean position has the maximum velocity and the extreme positions have the minimum or zero velocity. Also, try to keep the standard expressions related to an SHM in mind always as they are handy while dealing with such numerical problems.