Q. The equation of motion of a particle is $\frac{d^{2}y}{dt^{2}} + Ky = 0,$ where $K$ is positive constant. The time period of the motion is given by
A) $\frac{2\pi}{K}$
B) $2\pi K$
C) $\frac{2\pi}{\sqrt{K}}$
D) $2\pi\sqrt{K}$
Explanation
(c) On comparing with standard equation $\frac{d^{2}y}{dt^{2}} + \omega^{2}y = 0$
we get $\omega^{2} = K \Rightarrow \omega = \frac{2\pi}{T} = \sqrt{K} \Rightarrow T = \frac{2\pi}{\sqrt{K}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.