Answer
$1.64\;rad$
Work Step by Step
Let the two sinusoidal waves are moving in the positive direction of an $x$-axis
The mathematical form of the resultant displacement of the two waves is given by
$y(x, t)= y_m\sin(kx- \omega t)+y_m\sin(kx- \omega t+\phi)$
where $\phi$ is the phase difference between the two waves.
or, $y(x, t)= y_m\Big[\sin\frac{kx- \omega t+kx- \omega t+\phi}{2}\cos\frac{kx- \omega t-kx+ \omega t-\phi}{2}\Big]$
or, $y^{'}(x, t)=y_m\cos\frac{\phi}{2}\sin\Big(kx-\omega t+\frac{\phi}{2}\Big) ............(1)$
Comparing $(1)$ with $y^{'}(x, t)=(3.0\;mm)\sin(20x-4.0t+0.820\;rad)$, we obtain
$\frac{\phi}{2}=0.820\;rad$
or, $\phi=1.64\;rad$
Therefore, the phase difference of the two waves is $1.64\;rad$.
