The amplitude and phase of a wave formed by the superposition of two harmonic travelling waves, $$y_1(x, t) = 4\sin(kx - \omega t)$$ and $$y_2(x, t) = 2\sin(kx - \omega t + \frac{2\pi}{3})$$, are :
(Take the angular frequency of initial waves same
as $$\omega$$)

Let the two waves be
$$y_1(x,t)=4\sin\bigl(kx-\omega t\bigr)$$
$$y_2(x,t)=2\sin\bigl(kx-\omega t+\tfrac{2\pi}{3}\bigr)$$

Both waves have the same angular argument $$\theta=kx-\omega t$$, but different amplitudes and a phase difference $$\phi=\tfrac{2\pi}{3}$$.
Write each wave in the compact form $$A\sin\theta$$ and $$B\sin(\theta+\phi)$$, where
$$A=4,\; B=2,\; \phi=\tfrac{2\pi}{3}$$.

The superposition principle gives
$$y=y_1+y_2=A\sin\theta+B\sin(\theta+\phi)$$

For two sine functions with the same frequency, the resultant is again a sine function:
$$A\sin\theta+B\sin(\theta+\phi)=R\sin(\theta+\delta)$$
where

Amplitude formula:
$$R^2=A^2+B^2+2AB\cos\phi\quad -(1)$$

Phase formula:
$$\tan\delta=\frac{B\sin\phi}{A+B\cos\phi}\quad -(2)$$

Compute the trigonometric values of $$\phi=\tfrac{2\pi}{3}$$:
$$\cos\phi=\cos\!\left(\tfrac{2\pi}{3}\right)=-\tfrac12,\qquad\sin\phi=\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}$$

Substitute into $$(1)$$:
$$\begin{aligned}R^2&=4^2+2^2+2\cdot4\cdot2\cos\!\left(\tfrac{2\pi}{3}\right)\\&=16+4+16\left(-\tfrac12\right)\\&=20-8\\&=12\end{aligned}$$
$$\Rightarrow\;R=\sqrt{12}=2\sqrt3$$

Substitute into $$(2)$$:
$$\begin{aligned}\tan\delta&=\frac{2\cdot\left(\tfrac{\sqrt3}{2}\right)}{4+2\left(-\tfrac12\right)}\\&=\frac{\sqrt3}{4-1}\\&=\frac{\sqrt3}{3}=\frac1{\sqrt3}\end{aligned}$$
$$\therefore\;\delta=\tfrac{\pi}{6}$$ since $$\tan\left(\tfrac{\pi}{6}\right)=\tfrac1{\sqrt3}$$ and $$0\lt\delta\lt\pi$$.

Hence, the resultant wave is
$$y(x,t)=2\sqrt3\;\sin\!\bigl(kx-\omega t+\tfrac{\pi}{6}\bigr)$$

Amplitude  =  $$2\sqrt3$$, Phase  =  $$\tfrac{\pi}{6}$$.

The correct choice is Option D.