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)

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