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 interval during which it travels a distance A/2 starting from the extreme position is:

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

Answer 1

Answer: (D)

$\frac{3A}{T}$

Answer 2

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, $\frac{ A }{2}= A \cos \left(\frac{2 \pi}{ T } t '\right)$ $\Rightarrow \frac{2 \pi}{ T } t '=\frac{\pi}{3}$ $\Rightarrow t '=\frac{ T }{6}$ So, the required mean velocity over this time period $v _{ av }=\frac{\int\limits_{0}^{\frac{ T }{6}} v dt }{\int\limits_{0}^{\frac{ T }{6}} dt }$ $=\frac{6}{ T } \int\limits_{0}^{ T }- A \left(\frac{2 \pi}{ T }\right) \sin \left(\frac{2 \pi}{ T } t \right) dt$ $=\frac{6 \times 2 \pi A }{ T ^{2}} \times \frac{ T }{2 \pi}\left[-\cos \left(\frac{2 \pi}{ T } t \right)\right]_{0}^{\frac{ T }{6}}$ $=\frac{12 \pi A }{ T ^{2}}\left[\frac{1}{2}-1\right] \times \frac{ T }{2 \pi}$ $=-\frac{3 A }{ T }$ So, the magnitude of $v _{ av }$ is $\frac{3 A }{ T }$