The displacement of a particle executing S.H.M. is $x=a \sin (\omega t-\phi)$. Velocity of the particle at time $t=\frac{\phi}{\omega}$ is $(\cos 0^{\circ}=1)$.

Question

Aicharya · Accepted Answer

$x=a \sin (\omega t-\phi)$ The velocity $v$ is the first derivative of displacement $x$ with time $t$ : $v=\frac{d x}{d t}$ Let's compute the derivative: $\begin{aligned} & v=\frac{d}{d t}[a \sin (\omega t-\phi)] \ & v=a \cos (\omega t-\phi) \cdot \frac{d}{d t}(\omega t-\phi) \ & v=a \cos (\omega t-\phi) \cdot \omega \ & v=a \omega \cos (\omega t-\phi) \end{aligned}$ We need to calculate this velocity at time $t=\frac{\phi}{\omega}$ : $\begin{aligned} & v=a \omega \cos \left(\omega \cdot \frac{\phi}{\omega}-\phiight) \ & v=a \omega \cos (\phi-\phi) \ & v=a \omega \cos \left(0ight) \ & v=a \omega \cdot 1 \ & v=a \omega \end{aligned}$ Therefore, the correct answer is Option B: $a \omega$.

The displacement of a particle executing S.H.M. is $x=a \sin (\omega t-\phi)$. Velocity of the particle at…

Solution