Question

Q. A particle performs a linear S.H.M. of period 3 second. The time
taken by the particle to cover a distance equal to half the amplitude,
from the mean position is​

Answer 1

Considering that the SHM object starts from mean position we can say that it's equation of motion with respect to time is :

$x = A \sin( \omega t)$

Now as per the question , the object travels from mean position to half the amplitude , so x = A/2

$= > \dfrac{A}{2} = A \sin( \omega t)$

Cancelling the A term :

$= > \dfrac{1}{2} = \sin( \omega t)$

$= > \sin( \frac{\pi}{6} ) = \sin( \omega t)$

$= > \omega t = \dfrac{\pi}{6}$

Let time period be T , angular frequency be $\omega$ , hence we can say :

$= > ( \dfrac{2\pi}{T}) t = \dfrac{\pi}{6}$

$= > t = \dfrac{T}{12}$

Putting the value of T as 3 seconds :

$= > t = \dfrac{3}{12} = 0.25 \: sec$

So final answer is :

Time taken is 0.25 seconds.