Question

A particle executes linear S.H.M. with amplitude \(4\) cm. The magnitude of velocity and acceleration is equal when it is at \(3\) cm from mean position. Time period is

• A. \(\frac{3π}{\sqrt{2}}\) s
• B. \(\frac{6π}{\sqrt{7}}\) s
• C. \(\frac{2π}{\sqrt{7}}\) s
• D. \(\frac{4π}{\sqrt{7}}\) s

Hint:

Use v = omega sqrt(A^2 - x^2) and a = omega^2 x.

Answer 1

The Correct Option is B

Step 1: Ratio method:
$\dfrac va=\dfrac{\sqrt{A^2-x^2}}{\omega x}=1$.

Step 2: Solve for omega:
$\omega=\dfrac{\sqrt{A^2-x^2}}{x}=\dfrac{\sqrt7}3$.

Step 3: Period:
$T=\dfrac{2\pi}{\omega}=\dfrac{6\pi}{\sqrt7}$. Option (B).

Final Answer:
Setting v = a gives omega = sqrt 7 / 3. \[ \boxed{B} \]