Q 11-13-055JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
The displacement of a particle executing SHM is given by $x = 10\sin\left(\omega t + \dfrac{\pi}{3}\right)\ \text{m}$. The time period of motion is $3.14\ \text{s}$. The velocity of the particle at $t = 0$ is ______ m/s.
Numerical value type. Enter your answer.
Answer: 10
$\omega = \dfrac{2\pi}{T} = \dfrac{2\times3.14}{3.14} = 2\ \text{rad/s}$.
$$v = 10\omega\cos\left(\omega t + \frac{\pi}{3}\right) \Rightarrow v(0) = 10\times2\times\frac12 = 10\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics