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. `asqrt3`
C. `asqrt2`
D. `a`

Answer

D

Explanation

Path difference `(Delta x)=50cm=(1)/(2)m`
Phase difference `Delta phi=(2pi)/(lamda)xx Delta x implies phi=(2pi)/(1)xx(1)/(2)=pi`
Total phase difference `=pi-(pi)/(3)=(2pi)/(3)`
`impliesA=sqrt(a^2+a^2+2a^2cos(2pi//3))=a`