Equation of SHM - NEET Physics Questions
Question 41: easy

The differential equation of motion of a particle executing SHM is \(\frac{d^2y}{dt^2} + Ky = 0\) where \(K\) is positive constant. The time period of the oscillation is given by

1. \(\frac{2\pi}{K}\)
2. \(2\pi K\)
3. \(\frac{2\pi}{\sqrt{K}}\)
4. \(2\pi\sqrt{K}\)
View Answer

Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).