Question

A particle executes a simple harmonic motion of amplitude A. The distance from the mean position at which its kinetic energy is equal to its potential energy is:

• A. 0.91 A
• B. 0.71 A
• C. 0.81 A
• D. 0.51 A

Hint:

In simple harmonic motion, the distance from the mean position at which the kinetic energy equals the potential energy is always \( \frac{A}{\sqrt{2}} \), where \( A \) is the amplitude of the motion.

Answer 1

The Correct Option is B

In simple harmonic motion, the total mechanical energy is the sum of kinetic energy (\( KE \)) and potential energy (\( PE \)) at any point. The total energy in SHM is constant and is given by: \[ E = \frac{1}{2} m \omega^2 A^2, \] where: - \( m \) is the mass of the particle, - \( \omega \) is the angular frequency, - \( A \) is the amplitude. At any point, the kinetic energy \( KE \) is given by: \[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2), \] where \( x \) is the displacement from the mean position. The potential energy \( PE \) is given by: \[ PE = \frac{1}{2} m \omega^2 x^2. \]
Step 1: Equating Kinetic Energy and Potential Energy
We are given that the kinetic energy is equal to the potential energy. Therefore: \[ KE = PE, \] \[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2. \] Canceling common terms: \[ A^2 - x^2 = x^2, \] \[ A^2 = 2x^2, \] \[ x^2 = \frac{A^2}{2}. \] Thus, \[ x = \frac{A}{\sqrt{2}} \approx 0.707 A. \]
Step 2: Final Answer
The distance from the mean position at which the kinetic energy is equal to the potential energy is approximately \( 0.71 A \), which corresponds to option (B).