Two sources `S_1 " and "S_2` separated 2.0 m, vibrate according to equation `y_1= 0.03 sin pi t " and " y_2= 0.02 sin pi t " where "y_1, y_2` and t are in M.K.S unit. They send out waves of velocity 1.5 m/s. Calculate the amplitude of the resultant motion of the particle co-linear with `S_1 " and "S_2` and located at a point to the left of `S_1`.

Explanation

The situation is shown in figure 6.26.
The oscillations `y_(1)` and `y_(2)` have amplitudes `A_(1) = 0.03 m` and `A_(2) = 0.02 `m respectively.
The frequency of both sources is `n = (omega)/(2pi) = (1)/(2) = 0.5 Hz`
Now wavelength of each wave `lambda = (v)/(n) = (1.5)/(0.5) = 3.0 m`
The path differnece for all points `P_(2)` to the right of `S_(2) `is `Delta-(S_(1)P_(2) - S_(2)P_(2)) - S_(1)S_(2) = 2m` Phase difference `phi = (2pi)/(lambda) xx` path difference
= `(2pi)/(3) xx 2.0 = (4pi)/(3)`
The resultant amplitude for this point is given by `R = sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1) A_(2) cos phi)`
`sqrt((0.03)^(2) + (0.02)^(2) + 2 xx 0.03 xx 0.002 xx cos (4pi//3))`
Solving we get
`R = 0.0265 m`

The path difference for all points p, to the left of `S_(1)`
`Delta = (S_(2)p - S_(1)P) = S_(1)S_(2) = 2.0m`
Hence the resultant amplitude for all points to the of `S_(1)` is also 0.0265 m`
For a point Q midway between `S_(1)` and `S_(2)` the path difference is zero i.e. `phi = 0`. Hence constructive interferne takes place at Q thus amplitude at this point is maximum and given as
`R = sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1)A_(2))`