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

Step-by-step explanation:

Given

Potential Energy is given by

`U(x)=ax^2+(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-(2b)/(x^3)\right )`

i.e.
`ax=(b)/(x^3)`

`x_0=((b)/(a))^(0.25)`

So angular Frequency of small oscillation is given by

`\omega =\sqrt{(U''(x))/(m)}`

for
`m=1`

we get
`\omega =\sqrt{(U''(x_0))/(1)}`

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

`\omega =√(8a)`