Question

A particle performs S.H.M. from the mean position. Its amplitude is \( A \) and total energy is \( E \). At a particular instant its kinetic energy is \( \frac{3E}{4} \). The displacement of the particle at that instant is

• A. \( A \)
• B. \( \frac{A}{8} \)
• C. \( \frac{A}{4} \)
• D. \( \frac{A}{2} \)

Hint:

In S.H.M., the sum of the kinetic and potential energies is always constant. If you know the kinetic energy at a certain point, you can calculate the displacement using energy conservation.

Answer 1

The Correct Option is D

Step 1: Understanding the energy in S.H.M.
In simple harmonic motion (S.H.M.), the total mechanical energy is the sum of the kinetic energy and potential energy. The total energy \( E \) is constant and is given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( A \) is the amplitude. At any point during the motion, the kinetic energy is given by: \[ K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2) \] where \( x \) is the displacement from the mean position. Step 2: Using the given information.
It is given that the kinetic energy at a particular instant is \( \frac{3E}{4} \), so: \[ \frac{3}{4} E = \frac{1}{2} m \omega^2 (A^2 - x^2) \] Since \( E = \frac{1}{2} m \omega^2 A^2 \), substituting this in, we get: \[ \frac{3}{4} \times \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \] Simplifying: \[ \frac{3}{4} A^2 = A^2 - x^2 \] \[ x^2 = \frac{A^2}{4} \] \[ x = \frac{A}{2} \] Step 3: Conclusion.
Thus, the displacement of the particle at that instant is \( \frac{A}{2} \), which is option (D).