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: Use energy
$KE=\tfrac1{16}KE_{max}$, so $PE=\tfrac{15}{16}E$.

Step 2: Convert
$\dfrac{x^2}{A^2}=\dfrac{15}{16}$, so $x=\dfrac{\sqrt{15}}{4}A$, option (D).

Final Answer:
$x=\frac{\sqrt{15}}{4}A$, option (D). \[ \boxed{\dfrac{\sqrt{15}}{4}A} \]