Question

The displacement - time graph for a particle executing SHM is as shown in figure
Which of the following statement is correct?

• A. The velocity of the particle is maximum at $t=\frac{3}{4}T$.
• B. The velocity of the particle is maximum at $t=\frac{T}{2}$.
• C. The acceleration of the particle is maximum at $t=\frac{T}{4}$.
• D. The acceleration of the particle is maximum at $t=\frac{3}{4}T$..

Answer 1

The correct option is A The velocity of the particle is maximum at $t=\frac{3}{4}T$.

For the SHM,

The displacement of the particle is given by, $x=Acos\omega t$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left(Acos\omega t\right)=-A\omega sin\omega t$

Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}(-\omega Asin\omega t)=-{\omega }^{2}Acos\omega t$.

The corresponding velocity-time and acceleration time graph is as shown in the figure.

Option $A$