$\Delta \phi=\mathrm{k} \Delta \mathrm{x}$
$=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$
Hence destructive interference will occur at point $P.$
$A_{\text {resultant }}=\sqrt{(2 a)^{2}+a^{2}-4 a^{2}}=a$
It will remain constant with time.
${x_1} = a\sin (\omega \,t + {\phi _1})$, ${x_2} = a\sin \,(\omega \,t + {\phi _2})$
If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
$(A)$ $u=0.8 v$ 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 $f_5=1.5 f_0$