Exploring the Doppler Effect: Understanding Pitch Changes of Police Car Sirens

[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?

 

[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.

 

[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?

 

[HallsofIvy]

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, V_s is the speed of sound. V is the speed of the source (in this problem the police car) relative to you.

 

[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.

 

[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?

 

[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.

 

[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?

 

[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?

 

[dragonlau]

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

 

[Snazzy]

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

 

[dragonlau]

OK. I calculated two answers:
$$f=440\frac{343}{343-25} = 475 Hz$$
and:
$$f=440\frac{343}{343+25} = 410 Hz$$
For the question "calculate what the apparent frequency is when the car approaches you with a speed of 25 m/s", I think the first answer is correct,.. am I right?

But how can I determine how much the pitch will drop when the vehicle recedes at the same speed?