Physics - SHM Question with Solution | TestHub

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 $\omega =\frac{2\pi }{T}=\frac{2\pi }{4}=\frac{\pi }{2}$ 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:

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

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

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