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JEE MAIN 2022
27-06-2022 S2
Question
The equation of a particle executing simple harmonic motion is given by $x=\sin \pi\left(t+\frac{1}{3}\right) m$ At t = 1s, the speed of particle will be (Given : $\pi=3.14$ )
Select the correct option:
A
$0 \mathrm{~cm} \mathrm{~s}^{-1}$
B
$157 \mathrm{~cm} \mathrm{~s}^{-1}$
C
$272 \mathrm{~cm} \mathrm{~s}^{-1}$
D
$314 \mathrm{~cm} \mathrm{~s}^{-1}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & x=\sin \pi\left(t+\frac{1}{3}\right) \\ & x=\sin \left(\pi t+\frac{\pi}{3}\right) \\ & V=\frac{d x}{d t}=\cos \left(\pi t+\frac{\pi}{3}\right) \pi \\ & =-\pi \times \frac{1}{2}=157 \mathrm{~cm} / \mathrm{s}\end{aligned}$
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