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

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 $\frac{\pi }{3}$ 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

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

Question

A$2a$
B$a\sqrt 3 $
C$a\sqrt 2 $
D$a$

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