Question

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

\((a) \frac{\sqrt{5}}{2\pi}\)

\((b)\ \frac{4\pi}{\sqrt{5}}\)

\((c)\ \frac{2\pi}{\sqrt{3}}\) 

\((d)\ \frac{\sqrt{5}}{\pi}\)

Answer 1

Correct option is : \((b)\ \frac{4\pi}{5}\)

Amplitude A = 3 cm

when particle is at x = 2 cm

its |velocity| = |acceleration|

\(i.e.,\ \omega\sqrt{(A^2 - x^2)} = \omega^2x\)