Question

Two sinusoidal waves produced by two vibrating sources $A$ and $B$ of equal frequencies, are propagating to the point $P$ along a straight line. The amplitude of every wave at $P$ is ${}^{\prime }{a}^{\prime }$ and the phase of $A$ is ahead by $\frac{\pi }{3}$ than that of $B$. The distance $AP$ is greater than $BP$ by $50\text{cm}$. If the wavelength is $1\text{m},$ then the resultant amplitude at the point $P$ will be

• A. $2a$
• B. $a\sqrt{3}$
• C. $a\sqrt{2}$
• D. $a$

Answer 1

The correct option is D $a$

From the data given in the question,
Path difference $\left(\Delta x\right)=AP-BP=50\text{cm}\text{or}0.5\text{m}$
We know that,
Phase difference $\left(\Delta \varphi \right)=\frac{2\pi }{\lambda }\times \left(\Delta x\right)$
$\Rightarrow \Delta \varphi =\frac{2\pi }{1}\times \frac{1}{2}=\pi$
Due to path difference, phase of disturbance due to $A$ is ahead by $\pi$.

Given that, each wave has an amplitude of $a$ and phase of $\text{A}$ is ahead by $\frac{\pi }{3}$ than that of $\text{B}$.
$\therefore$ Total phase difference between waves due to $A$ and $B$
$\varphi =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$

Therefore, the amplitude of the resultant wave is given by
$R=\sqrt{{\left(A\right)}^2+{\left(B\right)}^2+2AB\cos \varphi }$
Substituting the data ,
$R=\sqrt{a^2+a^2+2a^2\cos \left(\frac{4\pi }{3}\right)}$
$R=\sqrt{{\left(a\right)}^2+{\left(a\right)}^2-{\left(a\right)}^2}=a$
Hence, option (d) is the correct answer.