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 meters.
Choose the wrong option.

• A. The speed of the particle is maximum at $x=2\text{m}$.
• B. The particle executes simple harmonic motion.
• C. The period of oscillation of the particle is $\frac{\pi }{5}s$.
• D. The particle does not execute simple harmonic motion.

Answer 1

The correct option is D The particle does not execute simple harmonic motion.

Given that,
mass of the particle $m=0.1\text{kg}$
Potential energy is $U=5x^2-20x$,
For conservative forces,
$F=-\frac{dU}{dx}=-(10x-20)$
$\Rightarrow F=-10(x-2)$
$\Rightarrow F=-10X$ (Assume $x-2=X$)
$\Rightarrow ma=-10X$ ($a=\text{acceleration of the particle})$
$\Rightarrow 0.1\text{kg}\times a=-10X$
$\Rightarrow 0.1\text{kg}\times a=-10X$
$\Rightarrow a=-100X$ ...(i)
$\Rightarrow a\propto -X$
Particle is performing simple harmonic motion.
Option b is correct.

Comparing equation (i) with simple harmonic motion
$a=-{\omega }^{2}x=-100x$
$\Rightarrow \omega =10$
Time period of the simple harmonic motion is
$T=\frac{2\pi }{\omega }=\frac{2\pi }{10}$
$T=\frac{\pi }{5}s$
option c is also correct.

At $x=2\text{m},F=0$ i.e. particle is at mean position and speed is maximum.
Option a is correct.

Why this question ? A conservative force at a point is negative of potential energy gradient at that position.