Two waves are propagating to the point $P$ by two sources $A$ and $B$ of equal frequency. The amplitude of…

Two waves are propagating to the point $P$ by two sources $A$ and $B$ of equal frequency. The amplitude of every wave at $P$ is $a$ and the phase of $A$ is ahead by $\frac{\pi}{3}$ than that of $B$ and the distance $AP$ is greater than $BP$ by 50 cm. If the wavelength is 1m, then the resultant amplitude at the point $P$ will be
  1. (a) 2a
  2. (b) $a\sqrt{3}$
  3. (c) $a\sqrt{2}$
  4. (d) a

Solution

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

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