Question

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=\frac{4}{3}\text{s}$ is

• A. $\frac{\sqrt{3}{\pi }^2}{32}{\text{cm/s}}^2$
• B. $-\frac{{\pi }^2}{32}{\text{cm/s}}^2$
• C. $\frac{{\pi }^2}{32}{\text{cm/s}}^2$
• D. $-\frac{\sqrt{3}{\pi }^2}{32}{\text{cm/s}}^2$

Answer 1

The correct option is D $-\frac{\sqrt{3}{\pi }^2}{32}{\text{cm/s}}^2$

From the $x-t$ graph of the particle,


At, $t=0$, the particle is at $x=0$.
Hence, the particle is starting $\text{SHM}$ from its mean position.
The equation of $\text{SHM}$ of particle will be,
$x=A\sin \omega t$ ...........(1)
From the graph, one oscillation has been completed in time, $T=8\text{s}$ and maximum displacement, $A=1\text{cm}$
So, angular frequency, $\omega =\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}\text{rad/s}$
On putting above data in (1)
$x=1\sin \left(\frac{\pi t}{4}\right)$
At, $t=\frac{4}{3}\text{s}$
$x=1\sin (\frac{\pi }{4}\times \frac{4}{3})=\frac{\sqrt{3}}{2}\text{cm}$
Now,
Acceleration of the particle at $t=\frac{4}{3}\text{s}$,
$a=-{\omega }^{2}x=-{\left(\frac{\pi }{4}\right)}^2\times \frac{\sqrt{3}}{2}$
$\Rightarrow a=-\frac{\sqrt{3}{\pi }^2}{32}{\text{cm/s}}^2$