Question

Two waves are propagating to the point $P$ along a straight line produced by two sources $A$ and $B$ which are simple harmonic in nature and of equal frequency. The amplitude of every wave at $P$ is $a$ and the phase of $A$ is $\frac{\pi }{3}$ ahead that of $B$ and the distance $AP$ is greater than $BP$ by $50cm$. Then, the resultant amplitude at that point $P$ will be $xa$ (wavelength is 1 meter) then the value of $x$ is

Answer 1

$\mathrm{Path}\mathrm{difference}\left(\Delta x\right)=50cm=\frac{1}{2}m$
$\therefore \mathrm{Phase}\mathrm{difference}\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$
$\Rightarrow \varphi =\frac{2\pi }{1}\times \frac{1}{2}=\pi$
Total phase difference =$\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
$\Rightarrow A=\sqrt{a^2+a^2+2a^2\cos (2\pi /3)}A=a$