Question

A particle executes linear S.H.M. with amplitude 4 cm. The magnitudes of velocity and acceleration are equal when it is 3 cm from the mean position. The time period of motion is

• A. \( \dfrac{6\pi}{\sqrt{7}} \)
• B. \( \dfrac{3\pi}{7} \)
• C. \( \dfrac{3\pi}{\sqrt{7}} \)
• D. \( \dfrac{6\pi}{7} \)

Hint:

Use velocity and acceleration expressions together when special conditions are given in SHM.

Answer 1

The Correct Option is A

Step 1: SHM equations.
Velocity at displacement \(x\):
\[ v = \omega \sqrt{A^2 - x^2} \] Acceleration at displacement \(x\):
\[ a = \omega^2 x \]
Step 2: Given condition.
\[ |v| = |a| \] \[ \omega \sqrt{A^2 - x^2} = \omega^2 x \]
Step 3: Substituting values.
\[ A = 4 \, \text{cm}, \quad x = 3 \, \text{cm} \] \[ \omega \sqrt{16 - 9} = \omega^2 \times 3 \] \[ \omega \sqrt{7} = 3\omega^2 \]
Step 4: Solving for angular frequency.
\[ \omega = \frac{\sqrt{7}}{3} \]
Step 5: Time period calculation.
\[ T = \frac{2\pi}{\omega} = \frac{2\pi \times 3}{\sqrt{7}} = \frac{6\pi}{\sqrt{7}} \]
Step 6: Conclusion.
The time period of motion is \( \dfrac{6\pi}{\sqrt{7}} \).