JEE Main 27 July 2021 Shift 2 Solved Paper

Section: Physics
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Question : 25 of 90
 
Marks: +1, -0
A particle executes simple harmonic motion represented by displacement function as
x(t)=A‌sin(ωt+ϕ)
If the position and velocity of the particle at t=0s are 2cm and 2ωcms−1 respectively, then its amplitude is x√2cm where the value of x is ____.
[27 Jul 2021 Shift 2]
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