Q85 / 90 · JEE Main 2021 (27 Jul Shift 2) · Physics

A particle executes simple harmonic motion represented by displacement function as $x (t) = A \sin (ω t + ϕ)$ . If the position and velocity of the particle at $t = 0 s$ are $2 cm$ and $2 ω cm s^{- 1}$ respectively, then its amplitude is $x \sqrt{2} cm$ where the value of $x$ is

Correct answer: 2

Solution

$x (t) = A \sin (ω t + ϕ)$

$v (t) = A ω \cos (ω t + ϕ)$

$2 = Asin ϕ$ $. . . (1)$

$2 ω = A ω \cos ϕ$ $. . . (2)$

From $(1)$ and $(2)$

$\tan ϕ = 1$

$ϕ = 45 °$

Putting the value of $ϕ$ in equation $(1)$

$2 = A \{\frac{1}{\sqrt{2}}\}$

$A = 2 \sqrt{2}$

$x = 2$

Hence, the correct answer is 2.