Doppler Effect?

You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 92 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 79 Hz. Take the speed of sound in air to be v = 340 m/s.

Find the numeric value, in hertz, for the frequency of the train whistle fs that you would hear if the train were not moving.

1 Answer
May 19, 2017

Answer:

#sf(85color(white)(x)Hz)#

Explanation:

As an estimate I would expect the frequency to be about halfway between the two values i.e #~=# 85.5 Hz.

We can show this as follows:

When the train is approaching we get:

#sf(f_(obs)=[(v)/(v-v_(source))]f_(source))#

Where v is the speed of sound.

We can find the speed of the train #sf(v_(source)#, from which we can get #sf(f_(source))#.

When the train is receding we get:

#sf(f_(obs)=[(v)/(v+v_(source))]f_(source))#

Putting in the numbers we get:

#sf(92=[(340)/(340-v_(source)]]f_(source)" "color(red)((1)))#

#sf(79=[(340)/(340+v_(source))]f_(source)" "color(red)((2)))#

Now divide #sf(color(red)((1))# by #sf(color(red)((2))rArr)#

#sf(92/79=([(340)/(340-v_(source))])/([(340)/(340+v_(source))])xxcancel(f_(source))/(cancel(f_(source)))#

#sf(92/79=[(cancel(340))/(340-v_(source))]xx[(340+v_(source))/(cancel(340))])#

#sf(1.1645=(340+v_(source))/(340-v_(source)))#

#sf(1.1645(340-v_(source))=(340+v_(source))#

#sf(395.95-1.1645v_(source)=(340+v_(source))#

#sf(2.1645v_(source)=395.95-340=55.949)#

#:.##sf(v_(source)=55.949/2.1645=25.85color(white)(x)"m/s")#

This is the speed of the train.

Now we can substitute this value back into #sf(color(red)((1))rArr)#

#sf(92=[(340)/(340-25.85)]f_(source))#

#sf(92=[1.0823]f_(source))#

#sf(f_(source)=92/1.0823=85color(white)(x)Hz)#

From the prediction this seems reasonable.