Doppler effect

1. Apr 9, 2008

dragonlau

When a police car passes you with a siren with a frequency of 440 Hz, a noticeable
drop in the pitch of the sound of the siren will be observed as the car passes by. But why the pitch suddenly drops when the car is passes by?

I am not pretty sure why I heard the pitch suddenly drops, but I think it is because the soundwave reaches his top or am I wrong?

2. Apr 9, 2008

Snazzy

When a source comes towards a stationary detector, the perceived frequency is given by:

$$f'=f\frac{V}{V-V_s}$$

When a source moves away from a stationary detector, the perceived frequency is given by:

$$f'=f\frac{V}{V+V_s}$$

You can see that the denominator is larger than the numerator in the 2nd equation when the source moves away from the detector, so the perceived frequency, or pitch, will be lower.

As the cop car moves towards you, the wavelengths get shorter, and so the frequency of the sound increases. As it moves away from you, the wavelengths get longer, and the frequency decreases.

3. Apr 10, 2008

dragonlau

Ah.. I get it.
So, I can use those formulas to calculate the apparent frequency? Imagine that the car approaching me with a speed of 25 m/s. For which can I replaced it with 25 m/s? And how can I calculate how much the pitch will drop when the car recedes at the same speed?
So, f' = 440 Hz or is it f = 440 Hz? I think I confused about which one is the apparent frequency, the f' or f?

4. Apr 10, 2008

HallsofIvy

Staff Emeritus
In
$$f'=f\frac{V}{V-V_s}$$
You were told that the perceived frequency is f'. f would be the frequency "at" the car so the relative velocity is V= 0. Although snazzy didn't say it explicitely, Vs is the speed of sound. V is the speed of the source (in this problem the police car) relative to you.

5. Apr 10, 2008

Snazzy

Actually, V would be the speed of sound (or the speed of the wave in any medium) and V_s would be the speed of the source. I would think the 440 Hz would be the actual frequency of the siren as well, not the perceived frequency but that's debatable.

Last edited: Apr 10, 2008
6. Apr 10, 2008

dragonlau

Alright.
I calculated the Hertz with these formulas.
For the first I have:
f = 440 Hz
Vb = 25 m/s
V = 343 m/s (speed of sound in air)
$$f=440\frac{343}{343-25} = 475 Hz$$
and second:
$$f=440\frac{343}{343+25} = 410 Hz$$
Honestly, which one is the apparent frequency and are my answers correcly?
Well, i am some confused about the freqency of 440 Hz, I assume this is the frequency of the source right?

Can I just subtract the first answer by the second answer to get to know how much the pitch will drop when the car recedes?

Last edited: Apr 10, 2008
7. Apr 10, 2008

Snazzy

I don't know what the question is. You just asked "why does the pitch drop when the car passes by?" I answered that question. What you're doing now is just reaffirming the answer with mathematics.

8. Apr 10, 2008

dragonlau

No, in post #3 I have posted another problem statement. Perhaps you misread this post?
In post #3 I need to calculate the 'apparent frequency' and how much the pitch will drop.
I hope I am clear enough?

9. Apr 10, 2008

Snazzy

Well, are you trying to calculate the frequencies of the siren before and after it reaches you, or are you calculating the frequency of the siren after it reaches you and comparing that answer with the actual frequency?

10. Apr 12, 2008

dragonlau

Well, I need to calculate the apparent frequency, but I don't know what apparent frequency is (what does it means?).

11. Apr 12, 2008

Snazzy

Apparent frequency is the frequency of the sound you hear due to the Doppler effect (f').

12. Apr 12, 2008

dragonlau

$$f=440\frac{343}{343-25} = 475 Hz$$
$$f=440\frac{343}{343+25} = 410 Hz$$