$ x = 2 \sin \omega t \,;$ $ y = 2 \sin \left( {\omega t + \frac{\pi }{4}} \right)$
The path of the particle will be :

${x}_{1}=5 \sin \left(2 \pi {t}+\frac{\pi}{4}\right)$ and ${x}_{2}=5 \sqrt{2}(\sin 2 \pi {t}+\cos 2 \pi {t})$
The amplitude of second motion is ....... times the amplitude in first motion.
($A$) 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
$\mathrm{y}=\mathrm{A}_{0}+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$
Then the amplitude of its oscillation is given by