Question

A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on thecoordinate ' x ' as U(x)=U'(1-cos(ax)); U' and a are constants. Find the period of small oscillations that the particle performs about the equilibrium position.

Answer 1

potential energy of the particle depends on the coordinate ‘x’ as $U=U'(1-cosax)$

where U' and a are constant terms.

we know, negative of rate for change of potential energy with respect to position of particle is given force.

i.e., $F=-\frac{dU}{dx}=-U'asinax$

and we know, if any particle of mass m, oscillates with angular frequency $\omega$ then,

force is given by, $F=-m\omega^2x$

so, $-m\omega^2x=-U'asinax$.....(1)

for small displacement , sinax ≈ ax

so, $-U'asinax\approx -U'a^2x$

now from equation (1), $-m\omega^2x=-U'a^2x$

or, $\omega=\sqrt{\frac{a^2U'}{m}}$

we also know, $\omega=\frac{2\pi}{T}$

so, time period, T = $2\pi\sqrt{\frac{m}{a^2U'}}$