Physics · 2022
JEE Main · 27 June 2022, Shift 2 · Q40
The equation of a particle executing simple harmonic motion is given by x= sin π(t+1/3) m. At t=1 s, the speed of particle will be (Given: π=3.14 ) -
The equation of a particle executing simple harmonic motion is given by $\displaystyle x=\sin \pi\left(t+\frac{1}{3}\right) m$. At $\displaystyle t=1 \mathrm{~s}$, the speed of particle will be
(Given: $\displaystyle \pi=3.14$ ) -
Official answer
(2)
$\displaystyle 157 \mathrm{~cm} \mathrm{~s}^{-1}$
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JEE Main 2022 Physics question, with the answer from NTA’s final answer key. Where our answers come from.