Question

A particle of mass 40 g executes simple harmonic motion with an amplitude of 2.0 cm. If the time period of oscillation is \( \frac{\pi}{20} \) s, the total mechanical energy of the system is:

• A. 128 J
• B. 128 mJ
• C. 12.8 mJ
• D. 256 mJ
• E. 2.56 mJ

Hint:

In SHM, total mechanical energy remains constant and depends on amplitude squared.

Answer 1

The Correct Option is C

The total energy in SHM is given by: \[ E = \frac{1}{2} k A^2 \] where \( k = m\omega^2 \) and \( \omega = \frac{2\pi}{T} \). Given \( T = \frac{\pi}{20} \): \[ \omega = \frac{2\pi}{\frac{\pi}{20}} = 40 \] \[ k = m\omega^2 = 0.04 \times 40^2 = 64 { N/m} \] Substituting \( A = 0.02 \) m: \[ E = \frac{1}{2} \times 64 \times (0.02)^2 = 0.0128 { J} = 12.8 { mJ} \]