Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The equation of motion of a particle is $\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} + \mathbf{K} y = 0,$ where K is positive constant. The time period of the motion is given by
$(a) \frac{2\pi}{K} (b) 2\pi K (c) \frac{2\pi}{\sqrt{K}} (d) 2\pi \sqrt{K}$
Text Solution
Verified by ExpertsThe correct answer is:
C
On comparing with standard equation $\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} + \omega^2 y = 0$ we get $\omega^2 = K \Rightarrow \omega = \frac{2\pi}{T} = \sqrt{K} \Rightarrow T = \frac{2\pi}{\sqrt{K}}$.
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