Two simple harmonic motions are represented by 
$$\begin{gathered} {\mathrm{y}}_1=5\left(\sin 2\mathrm{\pi t}+\sqrt{3} \cos 2\mathrm{\pi t} \right) \\ {\mathrm{y}}_2=5 \sin \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right) \end{gathered}$$
The ratio of the amplitude of two SHM's is:

Solution:

$$\begin{gathered} {\mathrm{y}}_1=5\times 2\left(\frac{1}{2}\sin 2\mathrm{\pi t} +\frac{\sqrt{3}}{2}\cos 2\mathrm{\pi t}\right) \\ =10\left(\sin 2\mathrm{\pi t} \cos \frac{\mathrm{\pi }}{3}+\cos 2\mathrm{\pi t} \sin \frac{\mathrm{\pi }}{3}\right) \\ {\mathrm{y}}_1=10 \sin \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{3}\right) \\ {\mathrm{y}}_2=5 \sin \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{10}{5}=\frac{2}{1} \end{gathered}$$