Question

A particle performs simple harmonic motion with a period of $2s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\left(\frac{1}{a}\right)s$. The value of ‘$a$’ to the nearest integer is _____.

Answer 1

Step 1. Given data:

Time period,$\left(T\right)=2sec.$

Time taken by the particle, $t=\frac{1}{a}$

Step 2. As we know that:

$x=A\sin (\omega t+\phi )$ [ Where, phase constant, $\phi =0$, at mean position, $\omega$ is angular velocity, $A$ is amplitude.]

$\frac{A}{2}=A\sin \left(\frac{2\pi }{T}\right)t$ [$\because T=\frac{2\mathrm{\pi }}{\omega }$and displacement equal to half of its amplitude.]

$\left(\frac{2\pi }{T}\right)t={\sin }^{-1}\left(\frac{1}{2}\right)$

$\left(\frac{2\pi }{T}\right)t=\frac{\pi }{6}$

$\left(\frac{2}{2}\right)t=\frac{1}{6}$ $\left[\because T=2sec\right]$

$\begin{array}{rcl}t& =& \frac{1}{6}\\ \frac{1}{a}& =& \frac{1}{6}\left[\because t=\frac{1}{a}\right]\end{array}$

$\therefore a=6$ .

Hence, the value of ‘$a$’ to the nearest integer is $\mathbf{6}$.