Two waves are propagating to the point P along a straight line pr… - Vidyadip

Question

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 $\frac{\pi }{3}$ 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

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Answer

(d) Path difference $(\Delta x)$ $ = 50\,cm\, = \frac{1}{2}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 = $\pi - \frac{\pi }{3} = \frac{{2\pi }}{3}$

==> $A = \sqrt {{a^2} + {a^2} + 2{a^2}\cos (2\pi /3)} = a$

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