The velocity-time ( v-t ) graph of a particle executing simple...

From the given information, the time taken to go from zero velocity to the next zero velocity is half of the time period.

$\dfrac{T}{2} = 2$ s $\Rightarrow T = 4$ s.

The angular frequency is $\omega = \dfrac{2\pi}{T} = \dfrac{\pi}{2}$ rad/s.

The maximum velocity is v_\max = A\omega.

$0.4\pi = A \left(\dfrac{\pi}{2}\right) \Rightarrow A = 0.8$ 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 is given by $a = -\omega^2 x$.

$a = -\left(\dfrac{\pi}{2}\right)^2 (0.8) = -\dfrac{\pi^2}{4} \times 0.8 = -0.2\pi^2$ m/s$^2$.

Alternatively, the velocity equation is $v(t) = 0.4\pi \sin\left(\dfrac{\pi}{2}t\right)$.

Acceleration $a(t) = \dfrac{dv}{dt} = 0.4\pi \left(\dfrac{\pi}{2}\right) \cos\left(\dfrac{\pi}{2}t\right) = 0.2\pi^2 \cos\left(\dfrac{\pi}{2}t\right)$.

At $t = 2$ s, $a(2) = 0.2\pi^2 \cos(\pi) = -0.2\pi^2$ m/s$^2$.

Answer: $-0.2\pi^2$ m/s$^2$