Question

$x-t$ graph of a particle undergoing SHM is shown. Acceleration of particle at $t=4/3s$ is

• A. $\sqrt{3}{\pi }^{2}m/s^2$
• B. ${\pi }^{2}m/s^2$
• C. $\frac{-{\pi }^2}{4}m/s^2$
• D. $-\sqrt{3}{\pi }^{2}m/s^2$

Answer 1

The correct option is D $-\sqrt{3}{\pi }^{2}m/s^2$

$x=-Asin\omega t$
$v=-A\omega cos\omega t$
& $a=A{\omega }^{2}sin\omega t$

$\because T=2s$
$\therefore \omega =\frac{2\pi }{T}=\pi$
& $A=2m$

$\therefore a$ at
$t=\frac{4}{3}s=2{\pi }^{2}sin\frac{4\pi }{3}m/s^2=-\sqrt{3}{\pi }^{2}m/s^2$