Question

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 } $

Answer 1

Answer: (C)

Velocity $=$ acceleration (Given)
$\therefore \omega \sqrt{a^{2}-y^{2}}=\omega^{2} y$
$\sqrt{(2)^{2}-(1)^{2}}=\omega(1)$
$\Rightarrow \omega=\sqrt{3}$
$T=\frac{2 \pi}{\omega}$
$\Rightarrow T=\frac{2 \pi}{\sqrt{3}}$