Amplitude of the first simple harmonic motion is

${\mathrm{A}}_1=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\sqrt{3}\mathrm{a}}{2}\right)}^2}=\mathrm{a}$

Amplitude of the second motion is

$\begin{array}{l}{\mathrm{A}}_2=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2}=\sqrt{2}\mathrm{a}\\ \therefore \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1}{\sqrt{2}}\end{array}$

We can rewrite the two motions as

$\begin{array}{c}{\mathrm{x}}_1=\mathrm{a}\left(\frac{1}{2}\sin \mathrm{\omega t}+\frac{\sqrt{3}}{2}\cos \mathrm{\omega t}\right)\\ =\mathrm{a}\left(\cos {60}^{\circ }\sin \mathrm{\omega t}+\sin {60}^{\circ }\cos \mathrm{\omega t}\right)\\ \mathrm{or} {\mathrm{x}}_1=\mathrm{asin}\left(\mathrm{\omega t}+{60}^{\circ }\right) ........\left(1\right)\\ \mathrm{and} {\mathrm{x}}_2=\frac{\mathrm{a}}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\sin \mathrm{\omega t}+\frac{1}{\sqrt{2}}\cos \mathrm{\omega t}\right)\\ =\frac{\mathrm{a}}{\sqrt{2}}\left(\cos {45}^{\circ }\sin \mathrm{\omega t}+\sin {45}^{\circ }\cos \mathrm{\omega t}\right)\\ \mathrm{or} {\mathrm{x}}_2=\frac{\mathrm{a}}{\sqrt{2}}\sin \left(\mathrm{\omega t}+{45}^{\circ }\right) ........\left(2\right)\end{array}$

From (1) and (2) it follows that the amplitudes are a and $\frac{a}{\sqrt{2}}$ (a result obtained above) the phases are 60 and 45°. Therefore, phase difference = 60°- 45° = 15°