The displacement time (x-t) graph of a particle performing simple...

From the given displacement-time graph, the amplitude of the simple harmonic motion is $A = 2$ m.

The time period of the motion is $T = 4$ s.

The angular frequency is given by $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$ rad/s.

The acceleration of a particle in SHM is given by $a = -\omega^2 x$.

At $t = 1$ s, the displacement of the particle is $x = +2$ m (as seen from the graph).

Substituting the values, we get:

$a = -\left(\dfrac{\pi}{2}\right)^2 \times 2$

$a = -\dfrac{\pi^2}{4} \times 2 = -\dfrac{\pi^2}{2}$ m/s$^2$.

Answer: $\dfrac{-\pi^2}{2}$ m/s$^2$