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 seconds 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)

$\frac{2\pi}{\sqrt{3}}$

Answer 2

Velocity $v = \omega\sqrt{A^{2} - x^{2}}$ and acceleration $= \omega^{2}x$

Now given, $\omega^{2}x = \omega\sqrt{A^{2} - x^{2}}$

⇒ $\omega^{2}.1 = \omega\sqrt{2^{2} - 1^{2}}$

⇒ $\omega = \sqrt{3}$ ∴ $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{3}}$