Question

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 $2 \, \omega \,cms ^{-1}$ respectively, then its amplitude is $x \sqrt{2} \, cm$ where the value of $x$ is ___

Answer 1

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