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)

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

Answer 2

For the given SHM,

The displacement of the particle is given by

$x = A\cos\omega t$

Velocity, $v = \frac{dx}{dt} = \frac{d}{dt}(A\cos\omega t) = - A\omega\sin\omega t$

Acceleration ,

$a = \frac{dv}{dt} = \frac{d}{dt}( - \omega A\sin\omega t) = - \omega^{2}A\cos\omega t$

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