Question

A particle is an oscillating simple harmonically with angular frequency $\omega$ and amplitude $A$. It is at a point (A) at a certain instant (shown in the figure). At this instant, it is moving towards mean position (B). It takes time $t$ to reach mean position (B). If the time period of oscillation is $T$, the average speed between $A$ and $B$ is :

• A. $\frac{A\sin \omega t}{t}$
• B. $\frac{A\cos \omega t}{t}$
• C. $\frac{A\sin \omega t}{T}$
• D. $\frac{A\cos \omega t}{T}$

Answer 1

The correct option is A $\frac{A\sin \omega t}{t}$

Let the S.h.M be $x=A\sin \omega t^{\prime }$, where $\omega =\frac{2\pi }{T}$

Velocity $v=\frac{dx}{dt}=\omega A\cos \omega t^{\prime }$

So, mean speed between $A$ to $B$ is $v_{av}=\frac{{\int }_t^0\omega A\cos \omega t^{\prime }d{t}^{\prime }}{{\int }_t^{0}d{t}^{\prime }}$

$\Rightarrow v_{av}=\frac{\omega A}{-t}.\frac{1}{\omega }{\left[\sin \omega t\right]}_t^0=\frac{-A\sin \omega t}{-t}$

$\Rightarrow v_{av}=\frac{A\sin \omega t}{t}$