$ \Rightarrow \frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} $
==> $\frac{3}{2} = \sqrt {\frac{{m + 2}}{m}} $
$ \Rightarrow \frac{9}{4} = \frac{{m + 2}}{m}$
$ \Rightarrow m = \frac{8}{5}kg = 1.6\;kg$
$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$

$y_1 =10 \sin \left(\omega t+\frac{\pi}{3}\right) cm$
$y_2 =5[\sin (\omega t)+\sqrt{3} \cos \omega t] \;cm$ respectively.
The amplitude of the resultant wave is $.............cm$.


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 ..... .