Q83 / 90 · JEE Main 2024 (04 Apr Shift 2) · Physics
The displacement of a particle executing SHM is given by $x=10 \sin \left(w t+\frac{\pi}{3}\right) m$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $t=0$ is ______$\mathrm{m} / \mathrm{s}$.
Correct answer: 10
Solution
Given,
$\begin{aligned}
& \mathrm{T}=3.14=\frac{2 \pi}{\omega} \\
& \omega=2 \mathrm{rad} / \mathrm{s} \\
& \mathrm{x}=10 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \text { at } \mathrm{t}=0 \\
& \mathrm{v}=10 \omega \cos \left(\frac{\pi}{3}\right)=10 \times 2 \times \frac{1}{2}[\mathrm{using} \omega=2 \mathrm{rad} / \mathrm{s}] \\
& \mathrm{v}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Hence, the correct answer is 10.