The displacement of a particle executing S.H.M. is $x=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$. Velocity of…
The displacement of a particle executing S.H.M. is $x=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$. Velocity of the particle at time $t=\frac{\phi}{\omega}$ is $\left(\cos 0^{\circ}=1\right)$
$\omega \cos \phi$
$\mathrm{a} \omega$
$\omega \cos 2 \phi$
$-\mathrm{a} \omega \cos 2 \phi$
Solution
Velocity of the particle is given as:
$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos (\omega \mathrm{t}-\phi)$
$\therefore \quad$ Velocity of the particle at time $\mathrm{t}=\phi / \omega$ is given as,
$\begin{aligned}
\quad \mathrm{v} & =\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos \left(\left(\omega \times \frac{\phi}{\omega}\right)-\phi\right)=\mathrm{a} \omega \cos 0 \\
\therefore \quad \mathrm{v} & =\mathrm{a} \omega
\end{aligned}$