Question

You are standing at a subway platform when a subway passes, sounding its whistle. If you observe the whistle to be at 12750Hz when it is approaching you and at 10750Hz after it passes,

Answer 1

As per doppler's Effect of sound we can say when subway is approaching the platform we will have

`f_1 = f_o* (v)/(v- v_s)`

`12750 = f_o* (340)/(340 - v_s)`

Similarly we can find the frequency when subway is passing away with same speed from us

`f_2 = f_o* (v)/(v + v_s)`

`10750 * f_o* (340)/(340 + v_s)`

now we can find the ratio of two

`(12750)/(10750) = (340 + v_s)/(340 - v_s)`

`1.186*(340 - v_s) = 340 + v_s`

`403.26 - 1.186 v_s = 340 + v_s`

`63.26 = 2.186 v_s`

`v_s = (63.26)/(2.186)`

`v_s = 28.94 m/s`

So the speed of subway is 28.94 m/s