Question

For a particle executing S.H.M., the potential energy is \(n\) times the kinetic energy when its displacement from mean position is \((\frac{2\sqrt{2}}{3})A\), where \(A\) is the amplitude of S.H.M. The value of \(n\) is

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

PE is half k x squared and KE is half k times (A squared minus x squared).

Answer 1

The Correct Option is D

Step 1: Approach
Use fractions of the total energy.

Step 2: Fraction of PE
PE is $\left(\dfrac xA\right)^2$ of the total: $\dfrac89$. So KE is $1-\dfrac89=\dfrac19$ of the total.

Step 3: Ratio
$n=\dfrac{8/9}{1/9}=8$. Option (D).

Final Answer:
At that position PE is 8/9 of the energy and KE is 1/9, so n = 8, option (D). \[ \boxed{8} \]