A train A crosses a station with a speed of 40 m/s and whitles a short pulse of natural frequency `n_0=596Hz` another train B is approaching towards the same station with the same speed along a parallel track, Two track are `d=99m` apart. When train A whistles. train B is 152 m away from the station as shown in Fig. If velocity of sound in air is `v=300(m)/(s)`. calculate frequency of the pulse heard by driver of train B.

Answer

724 Hz

Explanation

From figure id driver of train B receives the sound at point P, we use
`sqrt((99)^(2) +(152-x)^(2))/(330) = (x)/(40)` implies `(4)^(2)[(99)^(2) + (152)^(2) + (x)^(2) - 304 x] = (33)^(2)x^(2)`
`1073x^(2) + 4864x - 526480 = 0`
`x = (-4864 +sqrt((48640)^(2) + 4(1073)(526480)))/(2 xx 1073)` `(-4864 + 47784)/(2 xx 1073) = 20m`
Thus is figure
`cos theta = (132)/(sqrt((99)^(2) + (132)^(2))) = (132)/(165) = 0.8`
Apparent frequency is
`n_(ap) = 596[[330 + 40 xxcostheta]/[330 -cos theta]]`
`596[[330 + 32]/[330 - 32]]` = 724 Hz.