Q 11-13-102JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
A particle executes simple harmonic motion represented by displacement function as $x(t) = A\sin(\omega t + \phi)$. If the position and velocity of the particle at $t = 0$ s are $2$ cm and $2\omega\ \text{cm s}^{-1}$ respectively, then its amplitude is $x\sqrt2$ cm where the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
At $t = 0$: $A\sin\phi = 2$ and $A\omega\cos\phi = 2\omega \Rightarrow A\cos\phi = 2$.
$A^2 = 2^2 + 2^2 = 8 \Rightarrow A = 2\sqrt2$ cm, so $x = 2$.
Solution by Sreeraj P, M.Sc Physics