1. Let the two waves be $y_1 = a \sin \omega t, \qquad y_2 = a \sin(\omega t + \phi)$ where amplitude of each wave is $a = 8\text{ cm}$

2. The amplitude of the resultant of two waves of equal amplitude is $R = \sqrt{a^2 + a^2 + 2a^2\cos\phi}$ $R = a\sqrt{2+2\cos\phi}$

3. Given that the resultant amplitude is also $8$ cm, so $R = a$ Therefore, $a = a\sqrt{2+2\cos\phi}$

4. Dividing by $a$: $1 = \sqrt{2+2\cos\phi}$

5. Squaring both sides: $1 = 2+2\cos\phi$ $2\cos\phi = -1$ $\cos\phi = -\frac{1}{2}$

6. Hence, $\phi = 120^\circ$ (principal value; another equivalent value is $240^\circ$, but phase difference is usually taken as the smaller positive angle)

Therefore, the phase difference between the waves is: $\boxed{120^\circ}$