Two waves executing simple harmonic motion travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $$\sqrt{3}$$ times of amplitude of individual motions. The phase difference between the two motions is ______ (degree).

When two waves of the same amplitude $$A$$ and frequency are superimposed with a phase difference $$\phi$$, the resultant amplitude is given by:

$$A_R = \sqrt{A^2 + A^2 + 2A^2\cos\phi} = A\sqrt{2 + 2\cos\phi} = 2A\cos\left(\dfrac{\phi}{2}\right)$$

Given that the resultant amplitude equals $$\sqrt{3}A$$:

$$\sqrt{3}A = 2A\cos\left(\dfrac{\phi}{2}\right)$$

$$\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}$$

$$\dfrac{\phi}{2} = 30°$$

$$\phi = 60°$$

Therefore, the phase difference is $$\boxed{60}$$ degrees.