Question

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

Answer 1

Time period $\left(T\right)=2\text{s}$
Assume particle starts from its mean position ($t=0,x=0$), so its equation, $x=A\sin \left(\omega t\right)$

Let particle takes time $t_1$ to move from mean position to half of its amplitude.
$\frac{A}{2}=A\sin \left(\frac{2\pi }{T}\right)t$

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

$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{T}t$

$\Rightarrow t=\frac{1}{6}\text{s}=\frac{1}{a}\text{s}$

$\Rightarrow a=6$