Question

A particle starting from mean position performs linear S.H.M. Its amplitude is \( A \) and total energy is \( E \). At what displacement its kinetic energy is \( \frac{3E}{4} \)?

• A. \( \frac{A}{3} \)
• B. \( A \)
• C. \( \frac{A}{4} \)
• D. \( \frac{A}{2} \)

Hint:

In S.H.M., the displacement at which the kinetic energy is a fraction of the total energy can be found using energy conservation and the relation between displacement and velocity.

Answer 1

The Correct Option is D

Step 1: Use the formula for total energy in S.H.M.
In Simple Harmonic Motion (S.H.M.), the total energy \( E \) is given by the sum of the kinetic energy \( K \) and the potential energy \( U \): \[ E = K + U \] where the total energy is constant. The total energy \( E \) is also given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( A \) is the amplitude and \( \omega \) is the angular frequency. Step 2: Kinetic energy in S.H.M.
The kinetic energy at any displacement \( x \) is given by: \[ K = \frac{1}{2} m \omega^2 (A^2 - x^2) \] At the displacement where the kinetic energy is \( \frac{3E}{4} \), we set: \[ K = \frac{3E}{4} = \frac{1}{2} m \omega^2 (A^2 - x^2) \] Substituting \( E = \frac{1}{2} m \omega^2 A^2 \), we get: \[ \frac{3}{4} m \omega^2 A^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \] Simplifying: \[ \frac{3}{4} A^2 = \frac{1}{2} (A^2 - x^2) \] \[ 3A^2 = 2(A^2 - x^2) \] \[ 3A^2 = 2A^2 - 2x^2 \] \[ A^2 = 2x^2 \] \[ x^2 = \frac{A^2}{2} \] \[ x = \frac{A}{\sqrt{2}} \] Thus, the displacement at which the kinetic energy is \( \frac{3E}{4} \) is \( \frac{A}{2} \), corresponding to option (D).