Question

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

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

Answer 1

The correct option is D $\frac{4\sqrt{2}A}{T}$

Max. average velocity
$=\frac{\text{Total max displacement}}{\text{total time}}$

Particle will undergo maximum displacement when it passes through mean position midway during its time of travel $t=\frac{T}{4}$ as shown in figure below.



Hence, Maximum displacement is $\sqrt{2}A$

$V_{avg}=\frac{4\sqrt{2}}{T}A$