Two SHMs are represented by the equations ${\mathrm{y}}_1=10\sin (3\mathrm{\pi t}+\mathrm{\pi }/4)$ and ${\mathrm{y}}_2=5[\sin 3\mathrm{\pi t}+\sqrt{3}\cos 3\mathrm{\pi t}]$. Their amplitudes are in the ratio

$\begin{array}{l}{\mathrm{y}}_1=10\sin \left[3\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right]\\ {\mathrm{y}}_2=5[\sin (3\mathrm{\pi t})+\sqrt{3}\cos 3\mathrm{\pi t}]\\ =5\times 2\left[\frac{1}{2}\sin 3\mathrm{\pi t}+\frac{\sqrt{3}}{2}\cos 3\mathrm{\pi t}\right]\\ =10\left[\cos 60\cdot \sin 3\mathrm{\pi t}+\sin {60}^{\circ }\cdot \cos 3\mathrm{\pi t}\right]\\ =10[\cos \mathrm{B}\cdot \sin \mathrm{A}+\sin \mathrm{B}(\cos \mathrm{A}\left)\right]=10\sin (\mathrm{A}+\mathrm{B})\\ {\mathrm{y}}_2=10\sin \left(3\mathrm{\pi t}+{60}^{\circ }\right);{\mathrm{y}}_1=10\sin \left(3\mathrm{\pi t}+\frac{\mathrm{\pi }}{3}\right)\\ \text{ Ratio of amplitudes }=\frac{10}{10}=\frac{1}{1}\end{array}$