Question

A particle executes a linear S.H.M of amplitude 3 cm. When it is at 1 cm from the mean position, the magnitude of its velocity is half that of its acceleration at the point. The maximum velocity is

• A. \(\sqrt{2}\) cm/s
• B. \(4\) cm/s
• C. \(8\sqrt{2}\) cm/s
• D. \(12\sqrt{2}\) cm/s

Hint:

Equate velocity to half the acceleration at x = 1 cm to find omega, then use v max = A omega.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In SHM, \(v=\omega\sqrt{A^2-x^2}\), acceleration magnitude is \(\omega^2x\), and \(v_{max}=A\omega\).

Step 2: Use the condition
At \(x=1\text{ cm}\) with \(A=3\text{ cm}\): \(\omega\sqrt{9-1}=\dfrac{1}{2}\omega^2(1)\).

Step 3: Find omega
\(\omega=2\sqrt{8}=4\sqrt{2}\text{ rad/s}\)

Step 4: Maximum velocity
\(v_{max}=A\omega=3\times4\sqrt{2}=12\sqrt{2}\text{ cm/s}\)

Final Answer:
The maximum velocity is \(12\sqrt{2}\text{ cm/s}\). \[ \boxed{\text{(D) }12\sqrt{2}\text{ cm/s}} \]