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 `1` meter

Options

A. `a sqrt3`
B. `a sqrt2`
C. a
D. 2a

Answer

C

Explanation

Path difference `AP - BP = 50 cm = 0.5m`
`:.` Phase difference `= (2pi)/(lamda)` (path difference)
`= (2pi)/(1) xx 0.5 = pi`
`:.` Total phase difference `= pi - (pi)/(3) = (2pi)/(3)`
`:. a_(R) = sqrt(a^(2) + a^(2) + 2a .a cos ((2pi)/(3)))`
`= sqrt(2a^(2) + 2a^(2) xx (-(1)/(2)))`
`= sqrta^(2) = a`