Q. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :
A) $\frac{\sqrt{5}}{\pi}$
B) $\frac{\sqrt{5}}{2\pi}$
C) $\frac{4\pi}{\sqrt{5}}$
D) $\frac{2\pi}{\sqrt{3}}$
Explanation
[Solution] $\omega\sqrt{A^2 - x^2} = \omega^2x$ gives $\sqrt{5} = 2\omega$, so $\omega = \sqrt{5}/2$ and $T = 4\pi/\sqrt{5}$ s. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.