Question

Potential Energy (U) of a body of unit mass moving in a one-dimension conservative force field is given by,$U=(x^2-4x+3)$.All units in S.I (i) Find the equilibrium position of the body.(ii) Show that oscillations of the body about this equilibrium position is simple harmonic motion & find its time period.(iii) Find the amplitude of oscillations if speed of the body at equilibrium position is $2\sqrt{6}$m/s.

Answer 1

Potential Energy, $U=x^2-4x+3$
Now, the body is in one-dimension conservative force field so
$F=-\frac{dU}{dx}=-2x+4$
(i) At equilibrium $F=0-2x+4=0x=2m$
(ii) $F\left(x\right)=-2(x-2)$
But $F=ma$. Since m is constant and always positive,
$a\propto -(x-2)$
Therefore, this is simple harmonic motion.
Now, define new variable y such that $y=x-2$
$F\left(y\right)=-2y$
By comparing with $F\left(x\right)=-kx$
$k=2N/m$
Now, time period, $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{2}}=\sqrt{2}\pi$
(iii) At equilibrium position the velocity will be maximum which is:
$v=2\sqrt{6}=\omega A=\frac{2\pi }{T}A$
$A=\frac{vT}{2\pi }=\frac{2\sqrt{6}\sqrt{2}\pi }{2\pi }=2\sqrt{3}$