Question 16 of 194
The displacement of a particle executing SHM is given by . The time period of motion is . The velocity of the particle at is ______.
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Answer: 10
Given,
$\begin{aligned}
& \mathrm{T}=3.14=\frac{2 \pi}{\omega} \\
& \omega=2 \mathrm{rad} / \mathrm{s} \\
& \mathrm{x}=10 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \text { at } \mathrm{t}=0 \\
& \mathrm{v}=10 \omega \cos \left(\frac{\pi}{3}\right)=10 \times 2 \times \frac{1}{2}[\mathrm{using} \omega=2 \mathrm{rad} / \mathrm{s}] \\
& \mathrm{v}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\begin{aligned}
& \mathrm{T}=3.14=\frac{2 \pi}{\omega} \\
& \omega=2 \mathrm{rad} / \mathrm{s} \\
& \mathrm{x}=10 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \\
& \text { at } \mathrm{t}=0 \\
& \mathrm{v}=10 \omega \cos \left(\frac{\pi}{3}\right)=10 \times 2 \times \frac{1}{2}[\mathrm{using} \omega=2 \mathrm{rad} / \mathrm{s}] \\
& \mathrm{v}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$