Two simple harmonic motion are given by:

${\mathrm{x}}_1=\frac{\mathrm{a}}{2}\sin \mathrm{\omega t}+\frac{\sqrt{3}\mathrm{a}}{2}\cos \mathrm{\omega t}$ and ${\mathrm{x}}_2=\mathrm{asin} \mathrm{\omega t}+\mathrm{acos} \mathrm{\omega t}$, 

the minimum phase difference (in degree) between the two simple harmonic motions is xº. The value of x is _____.

We can rewrite the two motions as:

$$\begin{gathered} {\mathrm{x}}_1=\mathrm{a}\left(\frac{1}{2}\sin \mathrm{\omega t}+\frac{\sqrt{3}}{2}\cos \mathrm{\omega t}\right) \\ =\mathrm{a}\left(\cos {60}^{\circ }\sin \mathrm{\omega t} + \sin {60}^{\circ }\cos \mathrm{\omega t}\right) \end{gathered}$$

or   $x_1 = a \sin (\omega t+ 60°) \left(1\right)$
      $$\begin{gathered} x_2=\sqrt{2}a\left(\frac{1}{\sqrt{2}}\sin \omega t+\frac{1}{\sqrt{2}}\cos \omega t\right) \\ =\sqrt{2}a\left(\cos {45}^{\circ }\sin \omega t+\sin {45}^{\circ }\cos \omega t\right) \end{gathered}$$

or            $x_2=\sqrt{2}a\sin \left(\omega t+{45}^{\circ }\right) \left(2\right)$

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

 phase difference = 60°$-$45° = 15°