Discussion
The distance travelled by a particle executing linear S.H.M. from its mean position in 2s is equal to \(\frac{1}{\sqrt{2}}\) times its amplitude. Then its time period in seconds is
Answer: E
The time period is 16 s.
Starting from the mean position, \(x = A\sin\omega t\). After 2 s the displacement is \(\frac{A}{\sqrt{2}}\):
\[ \begin{aligned} \sin 2\omega &= \frac{1}{\sqrt{2}} \Rightarrow 2\omega = \frac{\pi}{4} \\ \omega &= \frac{\pi}{8},\quad T = \frac{2\pi}{\omega} = 16\ \text{s} \end{aligned} \]
The particle has moved through one eighth of a cycle (from 0 to \(\frac{\pi}{4}\) of phase), so 2 s is \(\frac{T}{8}\). 8 s would correspond to a quarter cycle, reaching the full amplitude.
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