Question

The potential energy of a particle of mass $0.1\text{kg}$, moving along the $x$-axis, is given by $U=5x(x–4)\text{J}$, where $x$ is in metres. It can be concluded that :

• A. The particle is acted upon by a constant force.
• B. The speed of the particle is maximum at $x=2\text{m}$.
• C. The particle executes simple harmonic motion.
• D. The period of oscillation of the particle is $\frac{\pi }{5}\text{s}$.

Answer 1

Answer: (B), (C), (D)

The period of oscillation of the particle is $\frac{\pi }{5}\text{s}$.

$U=5x(x-4)$
$\Rightarrow F=-\frac{dU}{dx}=-10x+20$
$\Rightarrow F=-10(x-2)$

This force is dependent on $x$, hence particle is acted upon by a variable force.

Since $F=-10x+20$ is of the form $F=-k(x-x_0)$, particle executes SHM.
Time period of oscillation
$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.1}{10}}=\frac{\pi }{5}\text{s}$

Speed is maximum when $a=\frac{F}{m}=0$ $\Rightarrow x=2\text{m}$

Hence, options (b), (c) & (d) are correct.