Answer
$\phi_2 = 2.7~rad$
Work Step by Step
Note that the phase difference between the two waves is $\phi_2-\phi_1$ which is $\phi_2$
We can find $\phi_2$:
$(2y_m) cos\frac{\phi_2}{2} = y_m'$
$cos\frac{\phi_2}{2} = \frac{y_m'}{2y_m}$
$\frac{\phi_2}{2} = cos^{-1}\frac{y_m'}{2y_m}$
$\frac{\phi_2}{2} = cos^{-1}\frac{4.0~mm}{(2)(9.0~mm)}$
$\frac{\phi_2}{2} = 1.35~rad$
$\phi_2 = 2.7~rad$
