Question

A particle is oscillating simple harmonically with angular frequency $\omega$ and amplitude $A$. It is at a point (A) at certain instant (shown in figure). At this instant it is moving towards mean positive (B). It takes time $t$ to reach position (B). If time period of oscillations 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 $\frac{A\sin \omega t}{t}$

$x=A\sin \omega t$

$dx=A\omega \cos \omega tdt$

$\left(v\right)=\frac{\int dx}{\int dt}$

$=\frac{{\int }_0^{t}A\omega \cos \omega tdt}{{\int }_0^{t}dt}$

$=\frac{A\omega \frac{\sin \omega t}{\omega }}{t}$

$=\frac{A\sin \omega t}{t}$.