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

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