$\Longrightarrow x=2 \cos \omega t=2 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left[\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]$
On comparing this with equation of $S H M: x=A \sin (\omega t+\phi)$
$\Longrightarrow A=2$ $\omega(\text { Angular frequency })$
$\Longrightarrow T=\frac{2 \pi}{\omega}$
$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .
