Question

Potential Energy $\left(U\right)$ of a body of unit mass moving in a one-dimension conservative force field is given by $U=(x^2-4x+3)$. All units are 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 and find its time period.
(iii) Find the amplitude of oscillations if the speed of the body at equilibrium position is $2\sqrt{6}m/s$.

Answer 1

Potential energy $u=(x^2-4x+3)J$

$1)$ At equilibrium position $\frac{du}{dx}=0$

So, $2x-4=0$

$\Rightarrow x=2m$

$2)$Force on the body $f=\frac{dv}{dx}=-2x-4$

$\Rightarrow f=-2(x-2)$

Let $u$ be displacement from mean position which is $x=2$

So, $x=u+2$

Hence $F=-2u$ It shows force acting on body is directly proportional to displacement of the body from mean position. So, it is executing S.H.M about $x=2$

$3)$ $F=-2u$

$\Rightarrow m\frac{d^{2}u}{{dt}^2}=-2u$, ($\frac{d^{2}u}{{dt}^2}$=acceleration of body from mean position.)

$\Rightarrow \frac{d^{2}u}{{dt}^2}+\frac{2}{m}.u=0$

So angular frequency $\omega =\sqrt{\frac{2}{m}}$

Velocity at mean position $\omega A=2\sqrt{6}$, ($A$=acceleration)

$\Rightarrow A=\frac{2\sqrt{6}}{\omega }$

$\Rightarrow A=\frac{2\sqrt{6}}{\sqrt{21}\sqrt{m}}$

$\Rightarrow A=2\sqrt{3}m$

Now, $m=1$, So, $A=2\sqrt{3}m$