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 `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

Options

A. 2a
B. `asqrt(3)`
C. `asqrt(2)`
D. a

Answer

D

Explanation

`Deltax` = path difference = 50 cm = `(1)/(2)` m
`Deltaphi=(2pi)/(lambda)xxDeltaximpliesDeltaphi=(2pi)/(1)xx(1)/(2)=pi`
Total phase difference `=pi-(pi)/(3)=(2pi)/(3)`
Resultant amplitude `A=sqrt(a^(2)+a^(2)+2a^(2)cos((2pi)/(3)))=a`