Question

A particle executing simple harmonic motion starts from mean position with amplitude '\(A\)' and periodic time '\(T\)'. At what displacement is its speed one-fourth of the maximum speed?

• A. \(\frac{A}{\sqrt{15}}\)
• B. \(\frac{A}{4}\)
• C. \(\frac{4A}{\sqrt{15}}\)
• D. \(\frac{A\sqrt{15}}{4}\)

Hint:

Use \(v=\omega\sqrt{A^2-x^2}\) and \(v_{max}=\omega A\).

Answer 1

The Correct Option is D

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

Step 2: Key Formula or Approach
We need \(v=\dfrac{\omega A}{4}\).

Step 3: Detailed Explanation
\[ \sqrt{A^2-x^2}=\frac A4 \Rightarrow A^2-x^2=\frac{A^2}{16} \]
\[ x^2=\frac{15A^2}{16}\Rightarrow x=\frac{A\sqrt{15}}{4} \]

Final Answer:
The displacement is \(\frac{A\sqrt{15}}{4}\), option (D). \[ \boxed{\dfrac{A\sqrt{15}}{4}\ \text{(D)}} \]