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 of shown in figure

The oscillations `y_1 " and "y_2` have amplitudes `A_1=0.03 m " and " A_2 = 0.02` respectively.
The frequency of both sources in `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 difference for all point P, to the left of `S_1`
`triangle = (S_2P-S_1P) = S_1S_2 = 2.0 m`
Hence the resultant amplitude for all points to the left of `S_1` is also 0.0265 m.