Q. A particle executes simple harmonic motion with an amplitude $4\ cm.$ At the mean position the velocity of the particle is $10\ cm/s.$ The distance of the particle from the mean position when its speed becomes $5\ cm/s$ is
A) $\sqrt{3}\ cm$
B) $\sqrt{5}\ cm$
C) $2(\sqrt{3})cm$
D) $2(\sqrt{5})cm$
Explanation
(c) $v_{\max} = a\omega \Rightarrow \omega = \frac{v_{\max}}{a} = \frac{10}{4}$
Now, $v = \omega\sqrt{a^{2} - y^{2}} \Rightarrow v^{2} = \omega^{2}\left( a^{2} - y^{2} \right) \Rightarrow y^{2} = a^{2} - \frac{v^{2}}{\omega^{2}}$
$\Rightarrow y = \sqrt{a^{2} - \frac{v^{2}}{\omega^{2}}} = \sqrt{4^{2} - \frac{5^{2}}{(10/4)^{2}}} = 2\sqrt{3}\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.