Question

The displacement-time graph for a particle executing $SHM$ is as shown in figure.

Which of the following statements 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

Answer: (A)

For the given $SHM$,
The displacement of the particle is given by, $x=Acosc\omega t$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left(Acos\omega t\right)=-A\omega sin\omega t$
Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}\left(-\omega Asin\omega t\right)=-{\omega }^{2}Acos\omega t$
The corresponding velocity-time and acceleration time graph is as shown in the figure.