Doppler Effect: How Does the Ambulance Siren Vary?

[annatar]

When an ambulance approaches, its siren sounds shriller than when I relatively motionless to it, and vice versa. Such is an example of doppler effect, isn't it?

However, there one thing I am not sure: does the siren gradually becomes more shriller when the ambulance approaches, or not?

(Assuming constant speed for the ambulance)

 

[Turv]

annatar,0

This is the equation for doppler effect of sound.

Example:

A sound source of 4000 hz is moving towards a stationary person at a speed of 80 m/s.

f = frequency at observer, c = speed of sound, V= velocity of source, f1= frequency of sound source.

f= f1 x c/ (c-v)

= 4000 x 340/ (340-80) = 5230 Hz

remember this will continue to vary.

 

[Doc Al]

However, there one thing I am not sure: does the siren gradually becomes more shriller when the ambulance approaches, or not?

If the ambulance comes directly at you, no, the observed frequency will be steady. But if it passes by you, then the observed frequency will vary from its highest to its lowest as it passes you. (What matters is the radial velocity of the siren with respect to you.)

 

[Turv]

Doc,

I'm confused, I've put up a good example of one scenerio, are you saying that on a US battle ship that the sailors will hear a steady frequency of Kamikazi pilots ? LOL!

 

[Doc Al]

Doc,

I'm confused, I've put up a good example of one scenerio, are you saying that on a US battle ship that the sailors will hear a steady frequency of Kamikazi Jap pilots ? LOL!

Read exactly what I wrote. If you are stationary in the water, and the plane is headed straight towards you, then yes, the observed frequency is steady.

Edit: It seems I misinterpreted your scenario with the planes. In the ambulance example, the speed of the ambulance was constant. (I thought you agreed; After all, you used a single speed in your calculation.) So I assumed you meant for the planes to have a constant speed as well, since you were comparing that example to the ambulance example. Of course, if the approaching sound source is accelerating, then the observed frequency will increase as the speed increases. (And as others have pointed out, it's likely that the planes are accelerating.) Sorry if I confused you.

 

[Turv]

You use the word steady? do you mean a frequency of 3000 HZ will remain 3000 Hz or steady means the frequency will build up constantly?

 

[Doc Al]

You use the word steady? do you mean a frequency of 3000 HZ will remain 3000 Hz or steady means the frequency will build up constantly?

By "steady" I mean unchanging. (Not steadily increasing!)

 

[annatar]

Hmm,so when the ambulance is coming/leaving, the sound heard will have a higher/lower frequency than the wave source, but remains steady at that frequency?

If the ambulance is accelerating directly towards you...I guess the frequency will rise then?

 

[Turv]

f = frequency at observer, c = speed of sound, V= velocity of source, f1= frequency of sound source.


Example 2:

A Jap plane heads towards a US battle ship:

f1 = 1500 Hz

V= 100 m/s

f= f1 x c/ (c-v)

f = 1500 x 340/ (340-100) = steady 1500 Hz

 

[Doc Al]

Hmm,so when the ambulance is coming/leaving, the sound heard will have a higher/lower frequency than the wave source, but remains steady at that frequency?

Yes. (If it's heading directly towards you at constant speed.)

If the ambulance is accelerating directly towards you...I guess the frequency will rise then?

Sure. The speed is increasing.

 

[annatar]

Thanks guys...I think I got it

 

[russ_watters]

Note that for an ambulance heading almost directly toward you, the velocity toward you is always decreasing and so the doppler shift is always changing (frequency is decreasing).

 

[Turv]

http://www.school-for-champions.com/science/sound_doppler_equations.htm"

Frequency

The equation for the observed frequency of sound when the source is traveling toward you is:

fo = fv/(v − vt)

where

* fo is the observed frequency
* f is the emitted frequency
* v is the velocity of sound
* vt is the velocity of the source toward you
* v > vt (vt is less than v)

Note that this equation does not work if the speed of the source is equal to the speed of sound. In such as case, you would be dividing by 0, which is impossible.

This matches my previous equations.

 

[nooma]

f = 1500 x 340/ (340-100) = steady 1500 Hz

so 1500 x 340/240 = 1500...
thats new...

 

[Vanadium 50]

Note that this equation does not work if the speed of the source is equal to the speed of sound. In such as case, you would be dividing by 0, which is impossible.

If an object is moving towards you at the speed of sound, you can't hear it until it's right on top of you. The equation is telling you something.

By the way "Jap" is considered an ethnic slur by many.

 

[Turv]

Vanadium,

You or others will not convince me until you put forward some calculations.

I've put mine forward with also a back up from that website i posted the link.

BTW I'm from the UK we aren't so frivolous here regarding "alleged slurs"

 

[Doc Al]

You or others will not convince me until you put forward some calculations.

Convince you of what? Is there something you don't understand about the Doppler effect?

 

[turin]

Turv, Kamikazes surely accelerate, since they are on a downward trajectory (I don't believe that they apply their brakes as they head for their target). So the pitch will increase due to their increased speed as they approach.

 

[russ_watters]

Vanadium,

You or others will not convince me until you put forward some calculations.

I've put mine forward with also a back up from that website i posted the link.

The ironic thing is that the calculation you posted doesn't imply what you concluded at the ende of your post! So if you don't want to believe your own math you'll have to go find some more math!

BTW I'm from the UK we aren't so frivolous here regarding "alleged slurs"

Now you know. Please incorporate this newfound undestanding into your future contributions to this site.

 

[russ_watters]

Turv, Kamikazes surely accelerate, since they are on a downward trajectory (I don't believe that they apply their brakes as they head for their target). So the pitch will increase due to their increased speed as they approach.

Good point!

It does include a couple of assumptions, that may or may not always be true:
1. It assumes that they are always on a downward trajectory (they probably usually are).
2. It assumes they haven't reached their terminal velocity.

 

[turin]

1. It assumes that they are always on a downward trajectory (they probably usually are).

I suppose. I find that a bit picky, though. Turv specifically asked about what would be heard on a ship. I don't think that the kamikazes were flying up out of the water. I suppose they could have skimmed the water. Anyway, the point is that the increase in pitch (as heard in the movies) occurs due to increased speed, not because the distance is decreasing.

2. It assumes they haven't reached their terminal velocity.

True.

 

[berkeman]

If an object is moving towards you at the speed of sound, you can't hear it until it's right on top of you. The equation is telling you something.

By the way "Jap" is considered an ethnic slur by many.

Good point. I've edited it out of post #4, and left it here as a reminder to all to please be aware of the terms we use.

 

[Doc Al]

You or others will not convince me until you put forward some calculations.

Please read the note I added to post number 5. (It wasn't clear to me that you were treating the ambulance and the planes as accelerating. That's certainly realistic for the attacking planes.)

 

[berkeman]

BTW I'm from the UK we aren't so frivolous here regarding "alleged slurs"

Fair enough, if true. However, the PF is US-based, hence the policy.

 

[DrGreg]

I'm from the UK we aren't so frivolous here regarding "alleged slurs"

Fair enough, if true.

Not true. I'm from the UK, too, and I found Turv's previous language just as inappropriate as everyone else did.

 

[Turv]

Gentlemen,

You don't just need to accelerate to achieve the doppler effect, it works with constant velocity too.

My formula is correct .

Drgreg, nice to see a fellow Brit on PF .

 

[turin]

Turv, I believe that the discussion of acceleration regards the kamikaze situation that you mentioned a while back. Radial acceleration of the source wrt the observer is required in order for the observed pitch to change. This may be a physical acceleration, as I suggest for the case of the kamikaze, or this may be an unphysical coordinate acceleration, as I suggest for the case of an ambulance that passes by at a constant speed. If the rate of change of the distance is constant, then the observed pitch is constant. If the rate of change of the distance changes, then the observed pitch changes.

Of course, I am making one assumption that was not pointed out by russ: the pitch produced by the source is constant, regardless of speed, which is probably quite unrealistic for kamikazes. Hmm ...

 

[Turv]

My dearest gentleman Turin,

higher pitch = higher frequency

see pitch scaling (Mels)

Thank you my dear Sir learned friend.

 

[turin]

higher pitch = higher frequency

I already know this. Was there something that I wrote that indicated otherwise? However, you do not know that I a gentleman.

 

[berkeman]

I already know this. Was there something that I wrote that indicated otherwise? However, you do not know that I a gentleman.

Settle down guys. I think the most recent confusion just came from this:

Of course, I am making one assumption that was not pointed out by russ: the pitch produced by the source is constant, regardless of speed, which is probably quite unrealistic for kamikazes. Hmm ...

For a given delta-V, there is a corresponding, constant Doppler shift. If the delta-V is changing (like the object is accelerating, like a plane in a dive), then the Doppler shift changes with the changing velocity.