Question

The distance travelled by a particle executing linear S.H.M. from its mean position in 2 s is equal to \( \frac{1}{\sqrt{2}} \) times its amplitude. Then its time period in seconds is

• A. \(10 \)
• B. \(8 \)
• C. \(9 \)
• D. \(12 \)
• E. \(16 \)

Hint:

Use \( x = A\sin(\omega t) \) for SHM position.

Answer 1

Answer: (E)

Concept: \[ x = A\sin(\omega t) \]

Step 1: Given.
\[ \frac{x}{A} = \frac{1}{\sqrt{2}} = \sin(\omega t) \] \[ \Rightarrow \omega t = \frac{\pi}{4} \]

Step 2: Substitute \( t=2 \).
\[ \omega = \frac{\pi}{8} \]

Step 3: Time period.
\[ T = \frac{2\pi}{\omega} = 16 \]