Question

A particle is executing simple harmonic motion with amplitude \( A \). At a distance \( x \) from the mean position, when the particle is moving towards the extreme position, it receives a blow in the direction of motion that instantaneously doubles its velocity. The new amplitude of the particle is:

• A. \( A \)
• B. \( \sqrt{A^2 - x^2} \)
• C. \( \sqrt{2A^2 - 3x^2} \)
• D. \( \sqrt{4A^2 - 3x^2} \)

Hint:

In SHM, when velocity doubles due to an external blow, the new amplitude can be derived using energy conservation principles.

Answer 1

The Correct Option is D

Step 1: Energy Conservation in SHM The total mechanical energy in simple harmonic motion is: \[ E = \frac{1}{2} m \omega^2 A^2 \] At displacement \( x \), the energy distribution is: \[ E_x = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m \omega^2 (A^2 - x^2) \] Since the velocity is doubled instantly, kinetic energy increases by a factor of 4: \[ \text{New total energy} = 4 \times \text{Original kinetic energy} + \text{Potential energy} \] Equating energies: \[ \frac{1}{2} m \omega^2 A_{\text{new}}^2 = \frac{1}{2} m \omega^2 (4A^2 - 3x^2) \] Solving for \( A_{\text{new}} \): \[ A_{\text{new}} = \sqrt{4A^2 - 3x^2} \] Conclusion Thus, the correct answer is: \[ \sqrt{4A^2 - 3x^2} \]