${x_1} = a\sin (\omega \,t + {\phi _1})$, ${x_2} = a\sin \,(\omega \,t + {\phi _2})$
If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
$\Rightarrow$ ${a^2} = {a^2} + {a^2} + 2{a^2}\cos \phi = 2{a^2}(1 + \cos \phi )$
$\Rightarrow$ $\cos \phi = - 1/2 = \cos 2\pi /3$ $\therefore$ $\phi = 2\pi /3$
$(A)$ $u=0.8 v$ and $f_5=f_0$
$(B)$ $u=0.8 v$ and $f_5=2 f_0$
$(C)$ $u=0.8 v$ and $f_5=0.5 f_0$
$(D)$ $u=0.5 v$ and $f_5=1.5 f_0$