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: Use energy fractions
$PE/E=x^2/A^2$ and total energy $E=PE+KE=9\,KE$.

Step 2: Solve
$PE=\tfrac89E$ so $\dfrac{x^2}{A^2}=\dfrac89$ and $x=\dfrac{2\sqrt2}{3}A$, option (C).

Final Answer:
$x=\frac{2\sqrt2}{3}A$, option (C). \[ \boxed{\dfrac{2\sqrt2}{3}A} \]