Question

A particle is doing SHM with maximum speed $v_0$, then maximum average speed of the particle for time interval $T/4$ is [$T$ is time period of SHM]

• A. $\frac{2\sqrt{2}{v}_0}{\pi }$
• B. $\frac{2v_0}{\pi }$
• C. $\frac{2\sqrt{2}{v}_0}{\pi }$
• D. $\frac{v_0}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{2\sqrt{2}{v}_0}{\pi }$

Let the equation of SHM be $x=Asin\left(\omega t\right)$
Then, velocity $v=A\omega cos\left(\omega t\right)$

It can be represented in the phasor diagram shown below.


From the phasor diagram, it is seen that the maximum average speed for a time interval of $\frac{T}{4}$ will be obtained for the part of the SHM marked in the phaser i.e. for a phase of ${45}^0$ on either side of the point of maximum velocity $v_0$ (that is at mean position)

Average velocity for the marked time interval $\frac{T}{4}$ is given by the expression

$v_0=A\omega \Rightarrow \omega =\frac{v_0}{A}T=\frac{2\pi }{\omega }<v>=\frac{\frac{2A}{\sqrt{2}}}{\frac{T}{4}}=\frac{8A}{T\sqrt{2}}<v>=\frac{2\sqrt{2}{v}_0}{\pi }$