The velocity-time ( v-t ) graph of a particle executing simple harmonic motion is a sine curve. The velocity of the particle is zero at t = 0 , reaches a positive maximum of 0.4 m/s at t = 1 s, and becomes zero again at t = 2 s. The instantaneous acceleration of the particle at t = 2 s is:

Step-by-step solution

From the given information, the time taken to go from zero velocity to the next zero velocity is half of the time period. T 2 = 2 s T = 4 s. The angular frequency is = 2 T = 2 rad/s. The maximum velocity is v_ max = A . 0.4 = A ( 2 ) A = 0.8 m. At t = 0 , the velocity is zero and subsequently becomes positive, which means the particle starts from the negative extreme position ( x = -A ). At t = 2 s, the particle has completed half a cycle and is at the positive extreme position, so x = +A = +0.8 m. The acceleration