Question

A particle of unit mass moves in a potential V(x)= ax^2+b/x^2 . Where a and b are constnts. Calcuate the angular frequency of small oscillations about the minimum of potential.

Answer 1

Answer:

Explanation:

Given

Potential Energy is given by

$U(x)=ax^2+\frac{b}{x^2}$

And Force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

Particle will be at equilibrium when Potential Energy is either minimum or maximum

$F=-\left ( 2ax-\frac{2b}{x^3}\right )$

i.e.$ax=\frac{b}{x^3}$

$x_0=(\frac{b}{a})^{0.25}$

So angular Frequency of small oscillation is given by

$\omega =\sqrt{\frac{U''(x)}{m}}$

for $m=1$

we get    $\omega =\sqrt{\frac{U''(x_0)}{1}}$

$U''(x_0)=2a+6a= 8a$

$\omega =\sqrt{8a}$