Question

A particle is performing simple harmonic motion. Equation of its motion is \( x = 5 \sin \left( 4t - \frac{\pi}{6} \right) \), \( x \) being the displacement from the mean position. Velocity (in ms\(^{-1}\)) of the particle at the instant when its displacement is 3, will be

• A. \( \frac{2\pi}{3} \)
• B. \( \frac{5\pi}{6} \)
• C. 20
• D. 16

Hint:

To find the velocity in SHM, differentiate the displacement equation and substitute the values at the given displacement.

Answer 1

The Correct Option is D

The displacement equation for simple harmonic motion (SHM) is: \[ x = A \sin(\omega t - \phi) \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Given: \[ x = 5 \sin \left( 4t - \frac{\pi}{6} \right) \] From the equation, we can identify that the amplitude \( A = 5 \) and the angular frequency \( \omega = 4 \, \text{rad/s} \). The velocity \( v \) is given by the derivative of the displacement with respect to time: \[ v = \frac{dx}{dt} = A \omega \cos(\omega t - \phi) \] At the instant when \( x = 3 \), we can find \( t \) by solving for \( \sin(\omega t - \phi) \): \[ 3 = 5 \sin \left( 4t - \frac{\pi}{6} \right) \] \[ \sin \left( 4t - \frac{\pi}{6} \right) = \frac{3}{5} \] Using this, we calculate the velocity as: \[ v = 5 \times 4 \times \cos \left( \arcsin \frac{3}{5} \right) \] which simplifies to: \[ v = 16 \, \text{ms}^{-1} \] Thus, the correct answer is (D).