Q. A particle of mass $m$ is located in a one dimensional potential field where potential energy is given by $(x) = A(1 - \cos{px)}$ , where $A$ and $p$ are constants. The period of small oscillations of the particle is
A) $2\pi\sqrt{\frac{m}{A\ p}}$
B) $2\pi\sqrt{\frac{m}{A\ p^{2}}}$
C) $2\pi\sqrt{\frac{m}{A}}$
D) $\frac{1}{2\pi}\sqrt{\frac{AR}{m}}$
Explanation
(b) $v_{x} = A(1 - \cos{px)}$
$F = - \frac{dv}{dx} = - Ap\sin{px}$
For small ($x$)
$F = - Ap^{2}x$
$a = - \frac{Ap^{2}}{m}x$
$a = \omega^{2}x$
$\omega = \sqrt{\frac{Ap^{2}}{m}}$
$\therefore$ $T = 2\pi\sqrt{\frac{m}{Ap^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.