Question

The displacement of a particle executing linear S.H.M. is \( x = 0.25 \sin (11t + 0.5) \) m. If \( \pi = \dfrac{22}{7} \), the period of S.H.M. is

• A. \( \dfrac{2}{7} \, s \)
• B. \( \dfrac{4}{7} \, s \)
• C. \( \dfrac{3}{7} \, s \)
• D. \( \dfrac{1}{7} \, s \)

Hint:

Always identify angular frequency directly from the SHM equation before calculating the time period.

Answer 1

The Correct Option is B

Step 1: Compare with standard SHM equation.
Standard equation of SHM is \[ x = A \sin(\omega t + \phi) \]
Step 2: Identify angular frequency.
Comparing, \[ \omega = 11 \, \text{rad s}^{-1} \]
Step 3: Write formula for time period.
\[ T = \frac{2\pi}{\omega} \]
Step 4: Substitute values.
\[ T = \frac{2 \times \frac{22}{7}}{11} = \frac{4}{7} \, s \]
Step 5: Conclusion.
The time period of the SHM is \( \dfrac{4}{7} \, s \).