Question

The displacement-time$(x-t)$ 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. $\frac{\sqrt{3}}{32}{\pi }^2{\text{cm/s}}^2$
• B. $\frac{-{\pi }^2}{32}{\text{cm/s}}^2$
• C. $\frac{{\pi }^2}{32}{\text{cm/s}}^2$
• D. $-\frac{\sqrt{3}}{32}{\pi }^2{\text{cm/s}}^2$

Answer 1

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

From given graph, amplitude $\left(A\right)=1\text{cm}$
Time period $\left(T\right)=8\text{s}$
$\therefore \omega =\frac{2\pi }{8}=\frac{\pi }{4}\text{rad/s}$
Dispalcement of the particle, $x=A\sin \omega t$
Acceleration , $a=-{\omega }^{2}A\text{sin}\omega t$
at $t=\frac{4}{3}\text{s}$,
$a=\frac{-{\pi }^2}{16}\times 1\times \text{sin}(\frac{\pi }{4}\times \frac{4}{3})$
$=\frac{-{\pi }^2}{16}\text{sin}\left(\frac{\pi }{3}\right)$
$=\frac{-\sqrt{3}}{32}{\pi }^2{\text{cm/s}}^2$