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, $v= \frac{dx}{dt } = \frac{d}{dt}\left(A cos \omega t\right) = -A\omega sin\omega t$
Acceleration, $a=\frac{dv}{dt} = \frac{d}{dt}\left(-\omega A sin\omega t\right) = -\omega^{2}A cos\omega t$
The corresponding velocity-time and acceleration time graph is as shown in the figure.