Q. The acceleration $\frac{d^{2}x}{dt^{2}}$ of a particle varies with displacement $x\ as\frac{d^{2}x}{dt^{2}}$ = $- kx$ where k is a constant of the motion. The time period T of the motion is equal to
A) $2\pi k$
B) $2\pi\sqrt{k}$
C) $2\pi\ /\sqrt{k}$
D) $2\pi/k$
Explanation
(c) $d^{2}x/dt^{2} = - kx;$
$T = 2\pi\sqrt{\frac{displacement}{acceleration}}$
So $T = 2\pi\sqrt{\frac{x}{kx}} = 2\pi\sqrt{\frac{1}{k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.