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:
The period \( T \) is \( T = 3.14 = \frac{2\pi}{\omega}. \)

Solving for \( \omega \):
The angular frequency is \( \omega = 2 \, \text{rad/s}. \)

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

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

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

Result: 10 m/s