Q. Two waves are propagating to the point $P$ along a straight line produced by two sources $A$ and $B$ of simple harmonic and of equal frequency. The amplitude of every wave at $P$ is $'a'$ and the phase of $A$ is ahead by $\pi/3$ than that of $B$ and the distance $AP$ is greater than $BP$ by $50\ cm.$ Then the resultant amplitude at the point $P$ will be, if the wavelength is $1\ meter$
A) $2a$
B) $a\sqrt{3}$
C) $a\sqrt{2}$
D) $a$
Explanation
(d) Path difference $(\Delta x) = 50\ cm = \frac{1}{2}m$
$\therefore$ Phase difference $\mathrm{\Delta}\phi = \frac{2\pi}{\lambda} \times \mathrm{\Delta}x \Rightarrow \phi = \frac{2\pi}{1} \times \frac{1}{2} = \pi$
Total phase difference $= \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\Rightarrow A = \sqrt{a^{2} + a^{2} + 2a^{2}\cos{(2\pi/3)}} = a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.