Question

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at a certain displacement '\(x\)' from the mean position. If '\(A\)' is the amplitude of S.H.M., the value of '\(x\)' is

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

Hint:

Use \(PE=\frac12m\omega^2x^2\) and \(KE=\frac12m\omega^2(A^2-x^2)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At displacement \(x\), \(PE=\tfrac12m\omega^2x^2\) and \(KE=\tfrac12m\omega^2(A^2-x^2)\).

Step 2: Key Formula or Approach
\(PE=8\,KE\) gives \(x^2=8(A^2-x^2)\).

Step 3: Detailed Explanation
\[ 9x^2=8A^2 \Rightarrow x=\frac{2\sqrt2}{3}A \]

Final Answer:
The displacement is \(\frac{2\sqrt2}{3}A\), option (C). \[ \boxed{\dfrac{2\sqrt2}{3}A\ \text{(C)}} \]