Question

A particle executes S.H.M. along a straight line with mean position $x = 0$, period $20s$ and amplitude $5cm$. The shortest time taken by the particle to go from $x = 4cm$ to $x = - 3cm$ is
A. $4s$
B. $7s$
C. $5s$
D. $6s$

Answer 1

Simple harmonic motion is a movement that is periodic which is back and forth through a mean position or central position so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete oscillation is always equal. The force accountable for the motion is always directed toward the mean position and is directly proportional to the distance from it.

Formula used:
$x = A(\sin \omega t + \phi )$ is used where $x$ is the distance of the particle from the mean position, $A$is the amplitude, $\omega$ is the angular frequency, $t$is the time taken and $\phi$ is the phase difference.
$\omega = \dfrac{{2\pi }}{t}$ is used where $\omega$ is the angular frequency, $t$ is the time taken.

Complete step by step answer:
It is given in the question that the time taken $t$ is equal to $20s$. Substituting this value in the formula $\omega = \dfrac{{2\pi }}{t}$, we get
$\omega = \dfrac{{2\pi }}{{20}} = \dfrac{\pi }{{10}}\dfrac{{rad}}{{\sec }}$
Let us consider at $t = 0$, $x$ will be equal to $4$.
$\Rightarrow 4 = 5\sin \phi$( $\because A = 5cm$)
$\Rightarrow {\sin ^{ - 1}}\dfrac{4}{5} = \phi$
Now at $t = {t_0}$, let $x$ be $- 3$. Hence we get
$\dfrac{{ - 3}}{5} = \sin (\omega {t_0} + \phi )$
$\Rightarrow {\sin ^{ - 1}}\dfrac{{ - 3}}{5} = \omega {t_0} + \phi$
$\Rightarrow {\sin ^{ - 1}}\dfrac{{ - 3}}{5} - {\sin ^{ - 1}}\dfrac{4}{5} = \omega {t_0}$
$\Rightarrow {t_0} = \dfrac{{10}}{\pi }({\sin ^{ - 1}}\dfrac{{ - 3}}{5} - {\sin ^{ - 1}}\dfrac{4}{5})$
On solving this, we get ${t_0} = 5s$

Hence,option (C) is the correct answer.

Additional Information:
Although simple harmonic motion is a simplification, it is still a wonderful approximation. Simple harmonic motion is vital in analysis to model oscillations like in wind turbines and vibrations in vehicle disbandments.

Note: Always remember that period and frequency are inverses of each other. They are inversely proportional with a coefficient of proportionality of one. Therefore, no coefficient is required to make their inverses equal. They are absolute and perfect reciprocals.