Question

The displacement of a particle varies with time as \( x = 4 \sin 3\omega t \). If its motion is simple harmonic, then its maximum acceleration is:

• A. \( 12 \omega^2 \)
• B. \( 36 \omega^2 \)
• C. \( 144 \omega^2 \)
• D. \( 24 \omega^2 \)

Hint:

The maximum acceleration in simple harmonic motion is proportional to the square of the angular frequency and amplitude.

Answer 1

The Correct Option is B

For simple harmonic motion, the maximum acceleration \( a_{{max}} \) is given by: \[ a_{{max}} = A \cdot \omega^2 \] Where: - \( A = 4 \) is the amplitude, - \( \omega = 3\omega \) is the angular frequency. 
Thus: \[ a_{{max}} = 4 \cdot (3\omega)^2 = 4 \cdot 9\omega^2 = 36 \omega^2 \] Final Answer: \( 36 \omega^2 \)