Question

The equation of motion of a particle executing simple harmonic motion is given by $ x = 0.6 \sin (3t + 0.8 \cos 5t) $, where $ x $ is in meters and $ t $ is in seconds. The maximum acceleration of the particle in m/s² is

• A. 9
• B. 12
• C. 18
• D. 6

Hint:

In SHM, the maximum acceleration is directly proportional to the square of the angular frequency and the amplitude.

Answer 1

The Correct Option is C

The maximum acceleration in SHM is given by \( a_{\text{max}} = \omega^2 A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude.
From the given equation, the angular frequency \( \omega = 3 \, \text{rad/s} \) and the amplitude \( A = 0.6 \).

Thus, the maximum acceleration is: \[ a_{\text{max}} = (3)^2 \times 0.6 = 18 \, \text{m/s}^2 \] Therefore, the maximum acceleration is 18 m/s².