Question

A particle execute SHM with time period $T$ and amplitude $A$. The maximum possible average velocity in time $\frac{T}{4}$:

• A. $\frac{2A}{T}$
• B. $\frac{4A}{T}$
• C. $\frac{8A}{T}$
• D. $\frac{4\sqrt{2}A}{T}$

Answer 1

Answer: (B)

$\frac{4A}{T}$

Given that,

Amplitude = $A$

Time period = $T$

Average velocity in time period = $\frac{T}{4}$

The displacement equation of SHM

$x=A\sin \omega t$

We know that,

$v=\frac{dx}{dt}$

$v=\frac{d\left(A\sin \omega t\right)}{dt}$

$v=A\omega \cos \omega t$

Now, the average velocity is

$<v{>}_{0\to \frac{T}{4}}=\frac{\underset{0}{\overset{\frac{T}{4}}{\int }}vdt}{\underset{0}{\overset{\frac{T}{4}}{\int }}dt}$

$<v>=\frac{\underset{0}{\overset{\frac{T}{4}}{\int }}A\omega \cos \omega t}{\frac{T}{4}}$

$<v>=\frac{4A\omega }{T}\underset{0}{\overset{\frac{T}{4}}{\int }}\cos \omega tdt$

$<v>=\frac{4A\omega }{T}{\left[\frac{\sin \omega t}{\omega }\right]}_0^{\frac{T}{4}}$

$<v>=\frac{4A}{T}[\sin \frac{2\pi }{T}\times \frac{T}{4}-0]$

$<v>=\frac{4A}{T}$

Hence, the average velocity is $\frac{4A}{T}$