Question

A particle is executing simple harmonic motion with a period of T seconds and amplitude A metre, The shortest time it takes to reach a point $\frac{A}{\sqrt{2}}m$ from its mean position in seconds is

Answer 1

Let at $t=0,x=0$

So, $x=A\sin \left(\omega t\right)$

$Atx=\frac{A}{\sqrt{2}}$

$\Rightarrow \frac{A}{\sqrt{2}}=A\sin \left(\omega t\right)$

$\Rightarrow \sin \left(\omega t\right)=\frac{1}{\sqrt{2}}$

For shortest time, $\omega t=\frac{n}{4}$

$\Rightarrow \left(\frac{2n}{T}\right)t=\frac{n}{4}$

$\Rightarrow t=\frac{T}{8}$