Question

A particle of mass \( m \) is executing simple harmonic motion about its mean position. If \( A \) is the amplitude and \( T \) is the period of S.H.M., then the total energy of the particle is

• A. \( \frac{4\pi^2 mA^2}{T^2} \)
• B. \( \frac{8\pi^2 mA^2}{T^2} \)
• C. \( \frac{2\pi^2 mA^2}{T^2} \)
• D. \( \frac{\pi^2 mA^2}{T^2} \)

Hint:

The total energy in simple harmonic motion depends on the mass, amplitude, and period. Use \( E = \frac{1}{2} m \omega^2 A^2 \) for energy calculations.

Answer 1

The Correct Option is C

Step 1: Understanding the total energy in S.H.M.
The total energy \( E \) in simple harmonic motion is given by the formula: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( \omega \) is the angular frequency and \( A \) is the amplitude. The angular frequency \( \omega \) is related to the period \( T \) by: \[ \omega = \frac{2\pi}{T} \] Step 2: Substituting \( \omega \) into the energy formula.
Substitute \( \omega = \frac{2\pi}{T} \) into the energy formula: \[ E = \frac{1}{2} m \left( \frac{2\pi}{T} \right)^2 A^2 = \frac{2\pi^2 m A^2}{T^2} \] Step 3: Conclusion.
Thus, the correct answer is (C), \( \frac{2\pi^2 mA^2}{T^2} \).