The instantaneous displacement of particle in S.H.M is $x=A \cos \left(\omega t+\frac{\pi}{4}\right)$. The…
The instantaneous displacement of particle in S.H.M is $x=A \cos \left(\omega t+\frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is
$\frac{\omega}{2 \pi}$
$\frac{\pi}{\omega}$
$\frac{2 \pi}{\omega}$
$\frac{\pi}{4 \omega}$
Solution
Given, $x=\operatorname{Acos}\left(\omega t+\frac{\pi}{4}\right)$
$\frac{\mathrm{dx}}{\mathrm{dt}}=(-\mathrm{A} \omega) \sin \left(\omega \mathrm{t}+\frac{\pi}{4}\right)$
When $\left(\omega t+\frac{\pi}{4}\right)=\frac{\pi}{2}, \sin \left(\omega t+\frac{\pi}{4}\right)=1$
$\therefore$ At time $t=\frac{\pi}{4 \omega}$, the velocity is maximum for the first time