Question

A small mass executes linear SHM about $O$ with amplitude $a$ and period, $T$. Its displacement from $O$ at time $\dfrac{T}{8}$ after passing through $O$ is
$A.\dfrac{a}{8}$
$B.\dfrac{a}{2\sqrt{2}}$
$C.\dfrac{a}{2}$
$D.\dfrac{a}{\sqrt{2}}$

Answer 1

In the process to solve the question we will apply the concept of simple harmonic motion. If a particle moves back and forth about a mean position in such a way that a restoring torque acts on the particle, which is proportional to displacement from mean position, but in the opposite direction from displacement, this motion of the particle is called simple harmonic motion. We will solve the above given question by applying the equation of displacement with amplitude for simple harmonic motion.

Formula used:
We are going to solve this problem by the use of the following relation:-
$x=a\sin (\omega t+\phi )$.

Complete step by step answer:
We will use the following relation of displacement and amplitude for simple harmonic motion to get the required solution:-
$x=a\sin (\omega t+\phi )$……………. $(i)$
Where $x$ is the displacement, $a$ is amplitude, $\omega$ is angular velocity, $t$ is any instant of time and $\phi$ is the phase shift which is measured in radians.
From the problem we have the following parameters with us:-
Amplitude is $a$, $t=\dfrac{T}{8}$, time period as $T$ and $\phi =0$ as it executes SHM from the mean position.
Now putting these parameters in $(i)$ we get
$x=a\sin \left( \omega \times \dfrac{T}{8}+0 \right)$
But we know that $\omega =\dfrac{2\pi }{T}$………….$(ii)$
Putting the value of $\omega$ is $(i)$ we get
$x=a\sin \left( \dfrac{2\pi }{T}\times \dfrac{T}{8} \right)$
$x=a\sin \left( \dfrac{\pi }{4} \right)$
$x=a\times \dfrac{1}{\sqrt{2}}$(as $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$)
$x=\dfrac{a}{\sqrt{2}}$.

So, the correct answer is “Option D”.

Note:
We have solved this problem by the application of relation of displacement and amplitude for a SHM. To find the velocity of a SHM we have different relations. Hence, don’t be confused in these two different conditions. The value of phase shift is taken as zero in this problem because of the fact that the particle executes SHM from the mean position. It is not necessary that $\phi =0$ in all cases.