1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Doppler Effect source/observer both moving

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A car and a bicycle are traveling directly toward each other at different speeds. The car horn has a frequency of 440Hz, and the car is traveling at 15.0m/s. The cyclist hears a sound of frequency 467Hz. The velocity of sound in air is 343m/s.

    i) Calculate the frequency of the signal emitted into the air by the moving car along the straight line joining the car to the bicycle.

    ii) Calculate the speed of the bicycle.

    iii) What frequency would the cyclist hear if the car and the bicycle were moving directly away from each other with the same speeds with which they were approaching each other in the previous part?

    2. Relevant equations

    For observer and source both moving:
    fobserver=fsource [(1+ or - vobserver/v)/(1+ or -vsource/v)]
    v= velocity of sound in air

    signs in numerator: + if observer moving toward source and - if away from source
    signs in denominator: - if source moving toward observer and + if away from observer


    3. The attempt at a solution

    For question i) it does not appear that I need to calculate anything. If the car horn has a frequency of 440 Hz then this is the frequency of the signal emitted into the air. If I am grossly missing something here please advise. It seems it is jusk asking for freq of the signal emitted into the air so for anything that would be whatever the frequency of the source is, right? Therefore, 440 Hz.

    ii) Using the equation above for source and observer moving toward each other:

    467=440[(1+vo/343)/1-15.0/343)]

    Solve for vo to get the speed of the bicycle: 4.5m/s

    iii) Using the equation above for source and observer moving away from each other:

    fobserver=440[(1-4.5/343)/1+15.0/343)] = 416 Hz

    Now this all seems pretty straightforward except that question i threw me off. Obviously, if I'm incorrect about my assumptions about frequency of the sound emitted into the air then the rest of my answers will also be wrong. If I am wrong, please help explain what fundamental concept about frequency I'm missing and how to then calculate the frequency of a sound emitted into air. Thanks!
     
    Last edited: Aug 4, 2012
  2. jcsd
  3. Aug 4, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The air horn emits frequency f to the air when it is stationary - but the horn has moved in the time the horn emits one wavelength... so the wavelength is shorter (ahead) when the source is moving than when it is stationary.

    The frequency transmitted to the air is the frequency heard by an observer stationary in the air ... you have a formula for that.
     
  4. Aug 5, 2012 #3
    Oh, of course! The way you explain it makes it much more clear. It is actually pretty obvious, so I am a little embarrassed I missed this. I will have no problem going ahead with the problem. Thank you for your time and help, Simon!
     
  5. Aug 5, 2012 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No worries - the question is being used to force you to realize where the Doppler effect physically comes from instead of just seeing it as an abstract set of formulae. This will help you figure out when you have something right or not.
     
  6. Aug 6, 2012 #5
    Thanks again, Simon. I am however, pondering something else now. I calculated the frequency of the signal emitted into the air using equation for a moving source and a stationary observer.

    However, would the remainder of the problem use the original f of source (ie: 440 Hz). I'm not sure, but I think, yes. My reasoning is that question two and three want to know velocity of the bike and frequency heard by cyclist if moving away from the car and the car moving away from the cyclist. Therefore, I would still use the calculations I worked out initially using the frequency of the car horn as 440 Hz because the equations are set up to accomodate for the changes for both source and observer moving. I hope that makes sense. Am on the right path? My book doesn't really explain this. It just indicates to use fs and doesn't indicate that the frequency of the moving source relative to a stationary observer needs to be figured initially and then used as fs.
     
  7. Aug 6, 2012 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You have to think about it in terms of where the Doppler effect is coming from.

    The car horn is creating waves in the air with a wavelength that is shorter in the direction of the bike than it would be if the car were stationary. Since the speed of sound hasn't changed - this means the frequency of the sound in the air (by v=fλ) is not the same as the frequency of vibrations of the horn.

    This is the frequency the cyclist would hear if she were stationary ... but she isn't: she's moving into the sound waves that are in the air. See?
     
  8. Aug 6, 2012 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Here - I'll walk you through Qi, to lead you into Qii, and leave Qiii up to you, using only the physics and not the equations. I'll translate the reasoning into math as I go.

    if c is the speed of sound, v is the speed of the car, u is the speed of the bike, and fH is the frequency of the horn on the car, then ...

    the horn emmits one wavelength in time period [itex]T=1/f_H[/itex].

    The "start" of each wave travels a distance cT in that time.
    When the horn emmits the "end" of the wave, it (and the car) has already travelled a distance vT. So: the end of the wave has hit the air a distance cT-vT behind the start of the wave ... the distance between the front and back of one wave is, by definition, the wavelength. So the wavelength of the sound in the air must be [tex]\lambda = (c-v)/f_H[/tex] ... and the frequency f heard by a stationary listener, in the path of the car, is given by:
    [tex]f = \frac{c}{c-v}f_H[/tex]
    ... confirm that this is the Doppler effect equation for a moving source :)

    Note: having left the horn, the sound moves independently of the car, through the air. If the car stopped moving, the compressed bit of sound would still remain compressed.

    But we don't want to know what a stationary listener hears - we want to know what the cyclist hears: which means we need to look at things from the point of view of the cyclist.

    The cyclist moves into the sound wave at speed u (wrt the ground remember), so, to the cyclist, each wavelength λ passes at speed c+u, so the frequency must be ...

    From this, you should be able to do Qiii as well.

    This is the closest I'll come to doing the problem for you.
    Wikipedia has a decent article on Doppler effect ... with animations and so on to help out.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook