Question

A person standing close to a railroad crossing hears the whistle of an approaching train. He notes that the pitch of the whistle drops as the train passes by and moves away from the crossing. The frequency of the approaching whistle is 524 Hz, and drops to 497 Hz after the train is well past the crossing. What is the speed of the train? Use 340 m/s for the speed of sound in air.

Answer 1

its the dobbler effect

Answer 2

Yes it is.

Answer 3

We know that the frequency of a wave is related to wavelength and velocity as such. \[f = {v \over \lambda}\]

Answer 4

"Dobbler" indeed.

I'm sorry, I just found that amusing.

Answer 5

its f=(v-V/v-V)*fo

Answer 6

v=340m/s but i am not sure what to do after that

Answer 7

As the train moves towards us, the relative velocity of the wave is increased. As it moves away, the relative velocity decreases. This creates a frequency shift.

Answer 8

so when it comes to us its v+V and when it goes away its v-V?

Answer 9

Slow down. Let's work through it.

Answer 10

ok

Answer 11

The doppler effect is expressed as\[f = \left ( c + v_r \over c + v_s \right)f_o\]In this case, we are standing still, therefore \(v_r = 0\). The doppler effect equation simplifies to \[f = \left ( c \over c + v_s \right) f_o\]where \(v_s\) is the velocity of the train.

Answer 12

We know the frequency as the train approaches and as it leaves, and we know the velocity of the sound wave. We also know that \(f_o\) is constant. We can write two expressions of the doppler effect equation. One as the train approaches and one as the train leaves.
Moving toward observer:\[f_1 = \left ( c \over c - v_s \right) f_o\]Moving away from observer:\[f_2 = \left ( c \over c + v_s \right ) f_o\]

Answer 13

Let's rearrange both equations to be in terms of \(f_o\). \[f_o = f_1 \left ( c - v_s \over c \right )\]and\[f_o = f_2 \left ( c + v_s \over c \right)\]
Now, let's set them equal to each other.

Answer 14

ok, is fo the initial? so 544?

Answer 15

*524

Answer 16

\[f_1 \left( c - v_s \over c \right) = f_2 \left( c+ v_s \over c \right)\]

Answer 17

\(f_1 = 524 Hz\) and \(f_2 = 497 Hz\).

Answer 18

We can solve that equation for \(v_s\)

Answer 19

can u explain to me what vs and vr mean?

Answer 20

\(v_s\) is the speed of the source. In this case, it is the speed of the train. \(v_r\) is the speed of the observer. In this case, \(v_r = 0\)

Answer 21

oh ok, i found vs to be 340 m/s :S?

Answer 22

oh wait hold on

Answer 23

Nope. If there were the case, f would be infinity as the source approached and we wouldn't hear anything.

Answer 24

ya cuz 340-340 i found an answer of 8.99m/s

Answer 25

\[f_1 \left ( 1 - {v_s \over c} \right) = f_2 \left( 1 + {v_s \over c} \right)\]\[f_1 - f_2 = f_1 {v_s \over c} + f_2 {v_s \over c}\]\[f_1 - f_2= {v_s \over c} (f_1 + f_2)\]\[v_s = \left (f_1 - f_2 \over f_1 + f_2 \right) c\]

Answer 26

yes i got it :D i think i kind of get the concept thank you so much :)

Answer 27

I also got 8.99 m/s.