JEE Main 2024 — 4 April, Shift 2 — Question 50

Given, $\mathrm{T}=3.14=\frac{2 \pi}{\omega}$

$\omega=2 \mathrm{rad} / \mathrm{s}$

$x=10 \sin \left(\omega t+\frac{\pi}{3}\right)$

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ at $\mathrm{t}=0$

$\mathrm{v}=10 \omega \cos \left(\frac{\pi}{3}\right)=10 \times 2 \times \frac{1}{2}[$ using $\omega=2 \mathrm{rad} / \mathrm{s}]$

$\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$