Question

A particle performs SHM along a straight line with the period $T$ and amplitude $A$. The mean velocity of the particle averaged over the time during which it travels a distance $A/2$ string from the extreme position is

• A. $\frac{A}{2T}$
• B. $\frac{A}{T}$
• C. $\frac{2A}{T}$
• D. $\frac{3A}{T}$

Answer 1

The correct option is D $\frac{3A}{T}$

The position of particle $x=A\cos \frac{2\pi }{T}t$ (As the particle starting from mean position)

Velocity $v=-A\frac{2\pi }{T}\sin \left(\frac{2\pi }{T}t\right)$

Let it will take t' time in reaching from $A$ to $\frac{A}{2}$.

So, $\frac{A}{2}=A\cos \left(\frac{2\pi }{T}{t}^{\prime }\right)$

$\Rightarrow \frac{2\pi }{T}{t}^{\prime }=\frac{\pi }{3}$

$\Rightarrow t^{\prime }=\frac{T}{6}$

So, the required mean velocity over this time period

$v_{av}=\frac{{\int }_0^{\frac{T}{6}}vdt}{{\int }_0^{\frac{T}{6}}dt}$

$=\frac{6}{T}{\int }_0^T-A\left(\frac{2\pi }{T}\right)\sin \left(\frac{2\pi }{T}t\right)dt$

$=\frac{6\times 2\pi A}{T^2}\times \frac{T}{2\pi }[-\cos \left(\frac{2\pi }{T}t\right){]}_0^{\frac{T}{6}}$

$=\frac{12\pi A}{T^2}[\frac{1}{2}-1]\times \frac{T}{2\pi }$

$=-\frac{3A}{T}$

So, the magnitude of $v_{av}$ is $\frac{3A}{T}$