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: Understanding the Concept:
In SHM with amplitude \(A\), at displacement \(x\): \(PE = \frac12kx^2\) and \(KE = \frac12k(A^2-x^2)\).

Step 2: Set up:
KE is \(25\%\) more than PE, so \(KE = 1.25\,PE = \frac54PE\).
\[ A^2 - x^2 = \frac54x^2 \Rightarrow A^2 = \frac94x^2 \]

Step 3: Solve:
\(x^2 = \frac49A^2\), so \(x = \frac23A = \frac23\times3 = 2\) cm.
Check: PE \(\propto 4\), KE \(\propto 9 - 4 = 5\), and \(5 = 1.25\times4\).

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