Question

A particle of mass m is located in a one dimensional potential field where potential energy is given by V (x) = A (l - cos px ), where A and p are constants. The period of small oscillations of the particle is

• A. $2\pi \sqrt{\frac{m}{Ap}}$
• B. $2\pi \sqrt{\frac{m}{ap^2}}$
• C. $2\pi \sqrt{\frac{m}{A}}$
• D. $\frac{1}{2\pi}\sqrt{\frac{Ap}{m}}$

Answer 1

Answer: (B)

$2\pi \sqrt{\frac{m}{ap^2}}$

Answer 2

We are given that a particle of mass m is located in a one dimensional potential field and the potential energy is given by V ( x) = A ( 1 - cos px ). So, we can find the force experiened by the particle as $\, \, \, \, \, \, \, \, F=-\frac{dV}{dx}=-Apsin \, px$ For small oscillations, we have $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, F \approx -Ap^2 x$ Hence, the acceleration would be given by $\, \, \, \, \, \, \, \, \, \, \, \, a=\frac{F}{m}=-\frac{Ap^2}{m}x$ Also we know that $\, \, \, \, \, \, \, \, \, \, \, \, a=\frac{F}{m}=-\omega^2 x$ So, $\, \, \, \, \, \, \, \, \, \, \, \, \omega =\sqrt{\frac{Ap^2}{m}}$ or $\, \, \, \, \, \, \, \, \, \, T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{ap^2}}$