Question

A particle performs S.H.M. with amplitude 'A'. The speed of the particle is \(\left(\frac{1}{3}\right)^{\text {rd }}\) of the maximum speed when its displacement from the mean position is

• A. \(\frac{\mathrm{A}}{3}\)
• B. \(\frac{\sqrt{2}\text{ }\mathrm{A}}{3}\)
• C. \(\frac{2\sqrt{2}\text{ }\mathrm{A}}{3}\)
• D. \(\frac{2\text{ }\mathrm{A}}{3}\)

Hint:

Use the speed formula of S.H.M. at displacement x and the maximum speed relation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In S.H.M. the speed at displacement \(x\) is \(v=\omega\sqrt{A^{2}-x^{2}}\) and the maximum speed is \(v_{max}=\omega A\).

Step 2: Set up
Given \(v=\dfrac{v_{max}}{3}\).
\(\omega\sqrt{A^{2}-x^{2}}=\dfrac{\omega A}{3}\)

Step 3: Calculation
Squaring: \(A^{2}-x^{2}=\dfrac{A^{2}}{9}\)
\(x^{2}=\dfrac{8A^{2}}{9}\), so \(x=\dfrac{2\sqrt{2}A}{3}\).

Step 4: Compare the options
This is option (C).

Final Answer:
The displacement is \(\dfrac{2\sqrt{2}A}{3}\). \[ \boxed{\text{(C) }\dfrac{2\sqrt{2}\,\mathrm{A}}{3}} \]