The displacement equations of two interfering waves are given by $$y_1 = 10 \sin\left(\omega t + \dfrac{\pi}{3}\right)$$ cm, $$y_2 = 5\left[\sin(\omega t) + \sqrt{3}\cos(\omega t)\right]$$ cm respectively. The amplitude of the resultant wave is ______ cm.

We need to find the amplitude of the resultant wave from two interfering waves.

Simplify $$y_2$$: $$y_2 = 5[\sin(\omega t) + \sqrt{3}\cos(\omega t)]$$

Using the identity $$a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2}\sin(\theta + \phi)$$ where $$\tan\phi = \frac{b}{a}$$:

$$\sin(\omega t) + \sqrt{3}\cos(\omega t) = 2\sin\left(\omega t + \frac{\pi}{3}\right)$$

(since $$\sqrt{1^2 + (\sqrt{3})^2} = 2$$ and $$\tan\phi = \sqrt{3}$$, so $$\phi = \frac{\pi}{3}$$)

$$y_2 = 5 \times 2\sin\left(\omega t + \frac{\pi}{3}\right) = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$

Compare with $$y_1$$: $$y_1 = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$

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

Both waves are identical — same amplitude, same frequency, same phase.

Find the resultant: $$y = y_1 + y_2 = 20\sin\left(\omega t + \frac{\pi}{3}\right)$$

The amplitude of the resultant wave is 20 cm.