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: Understanding the Concept:
In S.H.M. with angular frequency \(\omega\) and amplitude \(A\): velocity \(v=\omega\sqrt{A^2-x^2}\) and acceleration \(a=\omega^2x\) in magnitude.

Step 2: Equate them at x = 3 cm:
\[ \omega\sqrt{16-9}=\omega^2\cdot3 \]
\[ \sqrt7=3\omega\ \Rightarrow\ \omega=\frac{\sqrt7}3\ \text{rad/s} \]

Step 3: Period:
\[ T=\frac{2\pi}\omega=\frac{2\pi\cdot3}{\sqrt7}=\frac{6\pi}{\sqrt7}\ \text{s} \]

Step 4: Choose:
Option (B).

Final Answer:
The time period is 6 pi over sqrt 7 seconds. \[ \boxed{\frac{6\pi}{\sqrt7}\ \text{s}} \]