Q. A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is
A) $\frac{1}{2\pi\sqrt{3}}$
B) $2\pi\sqrt{3}$
C) $\frac{2\pi}{\sqrt{3}}$
D) $\frac{\sqrt{3}}{2\pi}$
Explanation
(c) Velocity = acceleration
$\omega\sqrt{a^{2} - y^{2}} = \omega^{2}y$
$\sqrt{{(2)}^{2} - {(1)}^{2}} = \omega(1)$
$\Rightarrow \ \ \ \ \ \ \ \ \ \omega = \sqrt{3}$
$T = \frac{2\pi}{\omega}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ T = \frac{2\pi}{\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.