1. Write the two waves in comparable form

Given: $y_1=10\sin\left(\omega t+\frac{\pi}{3}\right)\text{ cm}$ $y_2=5[\sin\omega t+\sqrt{3}\cos\omega t]\text{ cm}$

We first simplify $y_2$.

Recall: $\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$

Take $\phi=\frac{\pi}{3}$, then $\sin\left(\omega t+\frac{\pi}{3}\right)=\sin\omega t\cdot \frac{1}{2}+\cos\omega t\cdot \frac{\sqrt{3}}{2}$

Multiplying by $10$: $10\sin\left(\omega t+\frac{\pi}{3}\right)=5\sin\omega t+5\sqrt{3}\cos\omega t$

But $y_2=5\sin\omega t+5\sqrt{3}\cos\omega t$

So, $y_2=10\sin\left(\omega t+\frac{\pi}{3}\right)$

2. Compare $y_1$ and $y_2$

Now, $y_1=10\sin\left(\omega t+\frac{\pi}{3}\right)$ $y_2=10\sin\left(\omega t+\frac{\pi}{3}\right)$

Thus both waves are identical and are in the same phase.

3. Resultant amplitude

When two waves of amplitudes $A_1$ and $A_2$ are in phase, resultant amplitude is: $A=A_1+A_2$

Here, $A=10+10=20\text{ cm}$

4. Final answer

The amplitude of the resultant wave is: $\boxed{20\text{ cm}}$