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: Understanding the Concept
In SHM, potential energy at displacement \(x\) is \(\frac12kx^2\), and kinetic energy is \(\frac12k(A^2-x^2)\).

Step 2: Ratio
\[ n=\frac{PE}{KE}=\frac{x^2}{A^2-x^2} \]

Step 3: Substitute \(x=\dfrac{2\sqrt2}{3}A\)
\[ x^2=\frac{8}{9}A^2,\qquad A^2-x^2=\frac19A^2 \]
\[ n=\frac{8/9}{1/9}=8 \]

Step 4: Check the options
The value 2 or 4 would put the particle much nearer the mean position. At 8 the particle is close to the extreme, where PE dominates. So option (D).

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