Question

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

Answer 1

Answer: 2

$x(t)=A\sin (\omega t+\varphi )$
$v(t)=A\omega \cos (\omega t+\varphi )$
$2=A\sin \varphi \dots \dots .(1)$
$2\omega =A\omega \cos \varphi \dots \dots (2)$
From $(1)$ and $(2)\tan \varphi =1$ $\varphi =45^{\circ }$
Putting value of $\varphi$ in equation (1)
$2=A\left\{\frac{1}{\sqrt{2}}\right\}$
$A=2\sqrt{2}$
$x=2$