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

Path difference
$\Delta x=50\text{cm}=\frac{1}{2}m$
$A_a=A_b=a$
$\lambda =1m$
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$ = Phase difference
$\Rightarrow \Delta \varphi =\frac{2\pi }{1}\times \frac{1}{2}=\pi$
Total Phase difference $=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
Resultant amplitude
$R=\sqrt{a^2+a^2+2a^2\text{cos}\frac{2\pi }{3}}$
$=\sqrt{2a^2+2a^2(-\frac{1}{2})}$
$=\sqrt{2a^2-a^2}=\sqrt{a^2}=a$