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
Text Solution
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$Path difference (Δx) = 50 cm = \frac{1}{2} m$
$\therefore$ Phase difference $\Delta \varphi = \frac{2 \pi}{\lambda} \times \Delta x \Rightarrow \varphi = \frac{2 \pi}{1} \times \frac{1}{2} = \pi$ Total phase difference $= \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$
⇒ ⇒ $A = \sqrt{a^{2} + a^{2} + 2a^{2} \cos(2\pi/3)} = a$
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