Question

A $4 \,kg$ particle is moving along the $x$ -axis under the action of the force $F=-\left(\frac{\pi^{2}}{16}\right) x N .$ At $t=2 sec$, the particle passes through the origin and at $t=10 \,sec$ its speed is $4 \sqrt{2} m / s$. The amplitude of the motion is:

• A. $\frac{32 \sqrt{2}}{\pi} m$
• B. $\frac{16}{\pi} m$
• C. $\frac{4}{\pi} m$
• D. $\frac{16 \sqrt{2}}{\pi} m$

Answer 1

Answer: (A)

Acceleration of particle is
$a=-\left(\frac{\pi^{2}}{64}\right) x$
comparing with $a=-\omega^{2} x$ we get
$\Rightarrow \omega=\sqrt{\frac{\pi^{2}}{64}}=\frac{\pi}{8} $
$\Rightarrow T=\frac{2 \pi}{\omega}=16 \,s$
There is a time difference of $\frac{T}{2}$ between $t=2 s$ to $t=10 s$.
Hence particle is again passing through the mean position of SHM where its speed is maximum at $t=10 s$
$\Rightarrow V_{\max }=A \omega=4 \sqrt{2}$
$\Rightarrow A=\frac{4 \sqrt{2}}{\pi / 8}$
$=\frac{32 \sqrt{2}}{\pi} m .$