Question

The amplitude of a particle executing SHM is \(3\) cm. The displacement at which its kinetic energy will be \(25\%\) more than the potential energy is (in cm)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Use \(KE=\frac12k(A^2-x^2)\) and \(PE=\frac12kx^2\), with \(KE = 1.25\,PE\).

Answer 1

The Correct Option is B

Step 1: Plan:
Share the total energy between KE and PE.

Step 2: Steps:
Total energy $E = KE + PE = 1.25PE + PE = 2.25PE$, so $PE = \frac{E}{2.25} = \frac49E$.
Since $PE\propto x^2$ and $E\propto A^2$, $\frac{x^2}{A^2} = \frac49$, so $x = \frac23A = 2$ cm.

Final Answer:
The displacement is $2$ cm, option (B). \[ \boxed{2\ \text{cm}} \]