Question

A particle is executing SHM of periodic time $T$. The time taken by a particle in moving from mean position is half the maximum displacement is: $(\sin 30^{\circ}=0.5)$
A. $\dfrac{T}{2}$
B. $\dfrac{T}{4}$
C. $\dfrac{T}{8}$
D. $\dfrac{T}{12}$

Answer 1

Hint: Start by assuming the time, when the particle will be at the mean position .Then using the SHM equation for displacement of particle $x=A\sin(\omega t)$, find the time for the particle to undergo half maximum displacement.

Formula used: $x=A\sin(\omega t)$

Complete step-by-step answer:
SHM or simple harmonic motion is the motion caused by the restoring force; it is directly proportional to the displacement of the object from its mean position. And is always directed towards the mean.
The acceleration of the particle is given by, $a(t)=-\omega^{2}x(t)$
To find, the time taken by a particle in moving from mean position is half the maximum displacement
Let us assume that, at $t=0$, the particle is in the mean position.
Then from SHM, we know that the displacement of the particle is given by $x=A\sin(\omega t)$. Where, $A$ is the amplitude of oscillation and $\omega$ is the frequency of the oscillation
If, $t$ is the time taken for the particle to go from mean position to half maximum displacement. Then $\dfrac{A}{2} =A\sin(\omega t)$
$\sin(\omega t)=0.5$
Given that, $(\sin30^{\circ}=0.5)$
Then, $\sin(\omega t)= \sin 30^{\circ}=\sin\dfrac{\pi}{6}$
Then, $\omega t=\dfrac{\pi}{6}$
$\dfrac{2\pi}{T}t=\dfrac{\pi}{6}$
$t=\dfrac{T}{12}$
Hence, D. $\dfrac{T}{12}$ is the answer.

Note: Remember SHM motions are sinusoidal in nature. Assume, the particle is at mean when, $t=0$. This makes the further steps easier. Then, it will take time $t$ for the particle to move from a mean position to half maximum displacement.