Frequency of Sound in Wind: Firefighters and Sirens

In summary: I get this straight I act as if the wind is a observer??No. Nothing I have written remotely points to that. I meant that the speeds should be calculated wrt the air. Do you understand relative velocity? The wind is blowing means that the air medium as a whole is moving wrt the ground. Now imagine yourself stationary wrt the air. This means that you are moving wrt the ground, along with the air. If you consider the air to be still, then the ground and everything fixed to the ground will appear to be moving wrt you, in the opposite direction.Is it necessary to introduce all this complications here? It is not imperative, but makes the Physics easier to understand. Remember, the
  • #1
~christina~
Gold Member
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Homework Statement



A siren mounted on the roof of a firehouse emits sound at a frequency of 900Hz. A steady wind is blowing with a speed of 15.0m/s. Taking the speed of sound in calm air to be 343m/s, Find the wavelength of the sound

a) upwind of siren
b) downwind of siren

Firefighters are approaching the siren from different directions at 15.0m/s. What frequency does a firefighter hear

c) if she is approaching upwind position so that she is moving in the direction in which the wind is blowing and

d) if she is approaching from a downwind position and moving against the wind?


Homework Equations


Not sure but:

[tex] \lambda ' = \lambda - \Delta \lambda = \lambda - (v_s/ f) [/tex]

[tex] f'= v'/ \labmda= (v+ v_o)/ \lambda [/tex]

[tex] f'= ([v-v_o] / v)f [/tex] => observer moving away from source

[tex] f'= ([v+v_o] / v)f [/tex]= > observer moving toward source

[tex] f'= (v / [v-v_o] )f [/tex] => source moving toward observer

[tex] f'= (v / [v-v_o] )f [/tex]=> source moving away from observer

general...

[tex]f'= (v + v_o)/ (v-v_s) [/tex] => general doppler equation


The Attempt at a Solution



a) I'm not sure about exactly what "wind" is considered but I think I know that it affects sound waves. (confused about source and observer situation with wind blowing)

I think that if wind is blowing toward the firehouse where the siren is emmitting sound then it should decrease the wavelength of the sound wave going toward the direction the wind is blowing (upwind I think)

And if you are facing away from the wind (downwind?) then the wavelength would be larger right?

would I use these equations:
[tex] f'= (v / [v-v_o] )f [/tex] => source moving toward observer

[tex] f'= (v / [v-v_o] )f [/tex]=> source moving away from observer

then find f' and then use that to find the wavelength?
with: [tex] f'= v'/ \labmda= (v+ v_o)/ \lambda [/tex]

I need help on this.

Thanks
 
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  • #2
The formulae are all valid when the air is still, that is , the speeds are all wrt the air, so better make the wind go away by changing reference frame so that the air is still. (O=observer, S=source.)

Apply the general formula taking the proper sign into consideration in the num and denom. I hope you know how to do that. Go through some solved examples.

Let's do (a) first.

a) O is a fixed point upwind of S, which means wind is blowing from O toward S. In the frame of air, S is now moving in the air toward O, and O is moving wrt air away from S.

f’ = f(v +/- vo)/(v +/- vs), where v is the speed of sound.

How to put the correct signs now? If frequency increases due to movement of O, then the sign in the num should be +ve. Similarly, If frequency increases due to movement of S, then the sign in the num should be -ve.

O is moving wrt air away from S, so sign in num is -ve. S is moving wrt air toward O, so sign in denom is -ve.

After finishing this one, think about the others. Not too tough, really...:smile:
 
  • #3
Shooting Star said:
The formulae are all valid when the air is still, that is , the speeds are all wrt the air, so better make the wind go away by changing reference frame so that the air is still. (O=observer, S=source.)

Apply the general formula taking the proper sign into consideration in the num and denom. I hope you know how to do that. Go through some solved examples.

Let's do (a) first.

a) O is a fixed point upwind of S, which means wind is blowing from O toward S. In the frame of air, S is now moving in the air toward O, and O is moving wrt air away from S.

f’ = f(v +/- vo)/(v +/- vs), where v is the speed of sound.
I wasn't sure if this is what you meant but I drew a picture (arrow is supposed to be the other way)

http://img394.imageshack.us/img394/4648/83576799yq8.th.jpg

so if I get this straight I act as if the wind is a observer??

How to put the correct signs now? If frequency increases due to movement of O, then the sign in the num should be +ve. Similarly, If frequency increases due to movement of S, then the sign in the num should be -ve.

O is moving wrt air away from S, so sign in num is -ve. S is moving wrt air toward O, so sign in denom is -ve.

After finishing this one, think about the others. Not too tough, really...:smile:
 
Last edited by a moderator:
  • #4
~christina~ said:
so if I get this straight I act as if the wind is a observer??

No. Nothing I have written remotely points to that. I meant that the speeds should be calculated wrt the air.

Do you understand relative velocity? The wind is blowing means that the air medium as a whole is moving wrt the ground. Now imagine yourself stationary wrt the air. This means that you are moving wrt the ground, along with the air. If you consider the air to be still, then the ground and everything fixed to the ground will appear to be moving wrt you, in the opposite direction.

Is it necessary to introduce all this complications here? It is not imperative, but makes the Physics easier to understand. Remember, the formula for the change in frequency was derived when the medium was not moving. So, we pretend that the air is still and calculate all the speeds wrt air.

In your first post, you have written the formula for both change in [itex]\lambda[/itex] and f. Use only the formula for change in f. After finding the frequency as perceived by the observer, you can find the new [itex]\lambda[/itex]’ by f’[itex]\lambda[/itex]’= v+u, where u is the speed of wind relative to O and v is the speed of sound in still air, because that is the speed at which the sound waves come to the observer (in question a).

Now go back to question (a). If wind is blowing at speed u=15 m/s, then at what speed is the source S wrt air if you consider the air to be stationary? Same for the observer O. Here, O is taken to be an observer fixed to the ground.

Hint: After calculating, you should find f'=f.
 
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  • #5
Shooting Star said:
No. Nothing I have written remotely points to that. I meant that the speeds should be calculated wrt the air.

Do you understand relative velocity? The wind is blowing means that the air medium as a whole is moving wrt the ground. Now imagine yourself stationary wrt the air. This means that you are moving wrt the ground, along with the air. If you consider the air to be still, then the ground and everything fixed to the ground will appear to be moving wrt you, in the opposite direction.

I think I get it.
Is it necessary to introduce all this complications here? It is not imperative, but makes the Physics easier to understand. Remember, the formula for the change in frequency was derived when the medium was not moving. So, we pretend that the air is still and calculate all the speeds wrt air.

In your first post, you have written the formula for both change in [itex]\lambda[/itex] and f. Use only the formula for change in f. After finding the frequency as perceived by the observer, you can find the new [itex]\lambda[/itex]’ by f’[itex]\lambda[/itex]’= v+u, where u is the speed of wind relative to O and v is the speed of sound in still air, because that is the speed at which the sound waves come to the observer (in question a).
Now go back to question (a). If wind is blowing at speed u=15 m/s, then at what speed is the source S wrt ait if you consider the air to be stationary? Same for the observer O. Here, O is taken to be an observer fixed to the ground.

Hint: After calculating, you should find f'=f.

a) Um the source would be going at 15m/s if the wind is stationary since the air has no velocity when thought of this way. If an observer is fixed to the ground then the air would be going at 15m/s I think..but you said that f' would equal f and I'm confused about that since that wouldn't happen unless the answer to your question was that it was equal to 1 but vo= 0 so is that why f'=f?
But this would be when the observer is stationary, right? so that the ground is moving (source is not? (confusef becasuse you mention both

Thank you Shooting star
 
  • #6
~christina~ said:
a) Um the source would be going at 15m/s if the wind is stationary since the air has no velocity when thought of this way. If an observer is fixed to the ground then the air would be going at 15m/s I think..

That's right. So, vs=15 m/s wrt air. So is vo.

(EDITED: Away and toward got mixed up.)

Putting this in the general formula, f' = f(v-15)/(v-15) = f. The '-' in the numerator comes from the fact that the observer is moving away from the source in the air frame, and the '-' in the bottom comes from the fact that the source is moving toward the observer in the air frame.

but you said that f' would equal f and I'm confused about that since that wouldn't happen unless the answer to your question was that it was equal to 1 but vo= 0 so is that why f'=f?

Who said vo = 0? In the air frame vo has the value 15 m/s, because O is fixed to the ground in question a and b, and it is also moving wrt air.

Do this in your own way to understand properly. What has come out of all this is that if there is no relative speed between S and O, the frequency heard by O is the same as that emitted by S, even if the medium is moving wrt them. Not so for the wavelength.

Question b is almost the same. You can do it.
 
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  • #7
Shooting Star said:
That's right. So, vs=15 m/s wrt air. So is vo.

Putting this in the general formula, f' = f(v+15)/(v+15) = f. The + in the numerator comes from the fact that the observer is moving toward the source in the air frame, and the + in the bottom comes from the fact that the source is moving away from the observer in the air frame.

wow, I think I get it now.

Who said vo = 0? In the air frame vo has the value 15 m/s, because O is fixed to the ground in question a and b, and it is also moving wrt air.

Do this in your own way to understand properly. What has come out of all this is that if there is no relative speed between S and O, the frequency heard by O is the same as that emitted by S, even if the medium is moving wrt them. Not so for the wavelength.

I imagined myself as in a cloud and floating away from the firehouse with the siren.

Question b is almost the same. You can do it.
okay let's see if I really get it.

well downwind of the siren ...

would it be [tex] f'= f(v-v_o)/(v- vs) [/tex]?

I think it would be.
 
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  • #8
Shooting Star said:
What has come out of all this is that if there is no relative speed between S and O, the frequency heard by O is the same as that emitted by S, even if the medium is moving wrt them. Not so for the wavelength.

Have you noticed and understood this?

~christina~ said:
wow, I think I get it now.

well downwind of the siren ...

would it be [tex] f'= f(v-v_o)/(v- vs) [/tex]?

Remember, the air is being considered to be stationary, and vo and vs are the speeds of O and S respectively wrt air. What will be vo and vs? Are they equal? What about the + or - signs in the num and denom? Look at the solution for part (a).
 

Related to Frequency of Sound in Wind: Firefighters and Sirens

1. What is the relationship between wind and the frequency of sound?

The frequency of sound is affected by wind because wind is a physical force that can cause sound waves to either speed up or slow down. This change in speed leads to a change in the frequency of sound.

2. How does the frequency of sound impact firefighters and their ability to communicate?

The frequency of sound can impact firefighters by making it difficult for them to hear and communicate with each other. Wind can cause sounds to become distorted or muffled, making it challenging for firefighters to understand each other's commands and directions.

3. Can wind affect the frequency of sirens used by firefighters?

Yes, wind can affect the frequency of sirens used by firefighters. If the wind is blowing towards the siren, it can cause the sound waves to compress, resulting in a higher frequency. Similarly, if the wind is blowing away from the siren, it can cause the sound waves to stretch, resulting in a lower frequency.

4. How can firefighters adjust for wind when using sirens?

Firefighters can adjust for wind when using sirens by positioning the siren in a direction that is opposite to the wind. This will help to minimize any changes in frequency caused by the wind and ensure that the siren can be heard clearly by those nearby.

5. Are there any safety concerns for firefighters related to the frequency of sound in wind?

There are potential safety concerns for firefighters related to the frequency of sound in wind. If the wind is strong and the frequency of sound is significantly affected, it can make it difficult for firefighters to hear important warnings or instructions, potentially putting them in danger. It is important for firefighters to be aware of these potential risks and take necessary precautions when working in windy conditions.

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