Question

A particle executes simple harmonic oscillation with an amplitude $a$ . The period of oscillation is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
(A) $\dfrac{T}{4}$
(B) $\dfrac{T}{8}$
(C) $\dfrac{T}{{12}}$
(D) $\dfrac{T}{2}$

Answer 1

Use the formula of the simple harmonic motion given below and substitute the value of the displacement in it to find the value of the time taken. For the further simplification, the formula of the angular velocity is substituted to find the time taken.

Useful formula:
(1) The displacement equation of the particle that executes the simple harmonic motion is given as

$y = a\sin \omega t$

Where $y$ is the displacement of the particle, $a$ is the amplitude of the particle and the $\omega$ is the angular velocity and $t$ is the time taken.

(2) The angular velocity is given as

$\omega = \dfrac{{2\pi }}{T}$

Where $T$ is the time period.

Complete step by step solution:
Using the displacement equation,

$y = a\sin \omega t$

Since the time is calculated for the distance of half the amplitude, the displacement is taken as half the amplitude,

$\dfrac{a}{2} = a\sin \omega t$

By cancelling the similar terms in both sides of the equation, we get

$\sin \omega t = \dfrac{1}{2}$
$\omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

The value of the ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ is $\dfrac{\pi }{6}$ .

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

To find the value of the time taken,

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

Substituting the formula of the angular velocity in the above equation, we get

$t = \dfrac{\pi }{{6\dfrac{{2\pi }}{T}}}$

By simplifying the above equation, we get

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

Hence the time taken by the particle to travel the half of the amplitude is

$\dfrac{T}{{12}}$ .

Thus the option (C) is correct.

Note: The simple harmonic motion indicates the periodic motion in which the repetitive movement of the motion forth and the back, so that the displacement on one side is equal to the displacement of the other side. The time taken for the half of the amplitude is $\dfrac{T}{{12}}$ and that for the amplitude is $\dfrac{T}{6}$ .