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Siren and wind

  1. Mar 5, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data

    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?


    2. Relevant 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


    3. 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
     
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 5, 2008 #2

    Shooting Star

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    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:
     
  4. Mar 11, 2008 #3

    ~christina~

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    I wasn't sure if this is what you meant but I drew a picture (arrow is supposed to be the other way)

    [​IMG]

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

     
  5. Mar 11, 2008 #4

    Shooting Star

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    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.
     
    Last edited: Mar 11, 2008
  6. Mar 11, 2008 #5

    ~christina~

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    I think I get it.
    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
     
  7. Mar 11, 2008 #6

    Shooting Star

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    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.

    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.
     
    Last edited: Mar 11, 2008
  8. Mar 11, 2008 #7

    ~christina~

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    wow, I think I get it now.

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

    okay lets 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.
     
    Last edited: Mar 11, 2008
  9. Mar 12, 2008 #8

    Shooting Star

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    Have you noticed and understood this?

    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).
     
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