$\frac{\mathrm{A}}{2}=\mathrm{A} \sin \omega \mathrm{t}_{1} \quad \Rightarrow \omega \mathrm{t}_{1}=\frac{\pi}{6} \Rightarrow \mathrm{t}_{1}=\frac{\pi}{6 \omega}$
$\Rightarrow t_{1}=\frac{T}{12}$
and from $x=\frac{A}{2}$ to $x=A$ is
$\mathrm{t}_{2}=\frac{\mathrm{T}}{4}-\frac{\mathrm{T}}{12}=\frac{\mathrm{T}}{6} \Rightarrow \frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}=\frac{1}{2}$
$y_1 = \sin \left( {\omega t + \frac{\pi }{3}} \right)$ and $y_2 = \sin \omega t$ is :
($A$) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged
($B$) The final time period of oscillation in both the cases is same
($C$) The total energy decreases in both the cases
($D$) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases
$y_1 =10 \sin \left(\omega t+\frac{\pi}{3}\right) cm$
$y_2 =5[\sin (\omega t)+\sqrt{3} \cos \omega t] \;cm$ respectively.
The amplitude of the resultant wave is $.............cm$.

