Question

The displacement of a particle executing SHM is given by \( x = 10 \sin\left(\omega t + \frac{\pi}{3}\right) \, \text{m} \). The time period of motion is \( 3.14 \, \text{s} \). The velocity of the particle at \( t = 0 \) is ______ m/s.

Answer 1

Correct Answer: 10

Given:
\[ T = 3.14 = \frac{2\pi}{\omega}. \]

Solving for \( \omega \):
\[ \omega = 2 \, \text{rad/s}. \] 

The displacement \( x \) is given by:
\[ x = 10 \sin\left(\omega t + \frac{\pi}{3}\right). \] 

To find the velocity \( v \), differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = 10\omega \cos\left(\omega t + \frac{\pi}{3}\right). \] 

At \( t = 0 \):
\[ v = 10\omega \cos\left(\frac{\pi}{3}\right) = 10 \times 2 \times \frac{1}{2} = 10 \, \text{m/s}. \] 

Answer: 10 m/s