Question

The potential energy of a particle of mass $0.1kg$ moving along x-axis, is given by $U=5x(x-4)J$, where $x$ is in meters. It can be concluded that

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

Answer 1

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

The correct options are
B the speed of the particle is maximum at $x=2m$
C the particle executes simple harmonic motion
D the period of oscillation of the particle is $\pi /5s$

$F=-\frac{dU}{dx}=-\frac{d}{dx}(5x^2-20x)=-10x+20$
Here we see that force is directly proportional to the displacement and opposite to the direction of displacement. So it satisfies the condition of SHM. $\Rightarrow$ A is incorrect and C is correct.
Speed of the particle is maximum in mean position. i.e. where force is zero. i.e. $F=-10x+20=0\Rightarrow x=2m$
Speed of the particle is maximum at $x=2m$.. $\Rightarrow$ B is correct.
Since $F=-10x+20$ comparing it with general equation of SHM $F=-kx$, we have $k=10N/m$
$\Rightarrow T=2\pi \sqrt{\frac{m}{k}}$
$\Rightarrow T=2\pi \sqrt{\frac{0.1}{10}}=2\pi \sqrt{\frac{1}{100}}=\frac{2\pi }{10}=\frac{\pi }{5}s$ $\Rightarrow$ D is correct.