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The Doppler Effect - frequency of a bat reflected on wall

  1. Dec 6, 2011 #1
    The Doppler Effect -- frequency of a bat reflected on wall

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

    A bat flies toward a wall, emitting a steady sound with a frequency of 22.0 kHz. This bat hears its own sound plus the sound reflected by the wall.

    How fast should the bat fly to hear a beat frequency of 190 Hz?

    2. Relevant equations

    Doppler Effect equations

    3. The attempt at a solution

    I know this is a case of the Doppler effect, but I don't understand the variations of the equations in terms of when the source is stationary, listener is moving towards it, etc. I also understand that the frequency the bat emits will be different than the frequency rebounded by the wall. I kind of need to be walked through this one, because all of the explanations I've seen thus far aren't really getting through...

    Thanks!
     
  2. jcsd
  3. Dec 6, 2011 #2

    PeterO

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    Re: The Doppler Effect -- frequency of a bat reflected on wall

    you could try a first principles approach via wavelength.

    The reflected wave is to have a frequency 190 higher [it will be higher not lower as the bat is moving towards the wall]
    So the two frequencies detected are 22000 Hz and 22190 Hz

    calculate the wavelength of each of those waves.

    Lets assume they are 15 mm and 14 mm

    In order to achieve that apparent reduction, the bat would have to move forward 1mm between each cycle of sound produced.
    That means 1 mm every 22000th of a second. Scale that up and you will have a speed of 22 m/s. I doubt very much that the real wavelengths differ that much, so the real velocity will be less than that.
     
  4. Dec 8, 2011 #3
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    So can I just use the speed of sound in the air for my velocity in order to find the wavelength?
     
  5. Dec 8, 2011 #4
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    The solution shows use of the equation f[itex]_{L}[/itex]=[(v+v[itex]_{L}[/itex])/(v+v[itex]_{s}[/itex])]f[itex]_{s}[/itex]

    Could you perhaps explain it in terms of this equation? I understand that the positive direction will be from the listener to the source, so in the direction that the bat is flying. But I'm not entirely sure what I ought to be using for each variable.

    The solution doesn't show any use of wavelengths at all, so I'm a bit confused.
     
  6. Dec 8, 2011 #5

    PeterO

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    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Certainly that formula uses the speed of sound and the speed of "L" and "s" , where L is the listener and s is the source.

    The bat emits a sound - it is the source, the wall received the sound - it is a pseudo listener.

    The pitch of sound will arrive as a higher pitch.

    That sound is then reflected: the wall becomes the source and the bat the listener.

    The bat will detect that reflected sound [already a higher pitch that 22000] as an even higher pitch.

    Looks like you analyse it twice - and must have been given a speed of sound to use??



    My suggestion of using wavelengths was, as stated, to use first principles. An analysis of a general example will lead to a velocity expression - like this one shown here.
     
  7. Dec 8, 2011 #6
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Meh, ignore this post, I misread the initial problem. Sorry!
     
  8. Dec 8, 2011 #7
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    It seems I'm still doing something wrong. If I observe the instance in which the wall is the listener, I have f[itex]_{w}[/itex]=(v+v[itex]_{w}[/itex])/v-v[itex]_{b}[/itex])f[itex]_{b}[/itex]. Plugging in, 11700=(344/344-v[itex]_{b}[/itex])(1700). Which gives me a velocity of 294.02m/s, given that v[itex]_{b}[/itex] is my only unknown. But that means I've completely neglected the situation in which the bat is the listener and the wall is the source. I don't really see how to combine the two.

    For that matter, I'm confused by the equations I'm supposed to use and what variables they entail.

    For f heard by the wall I know it's [(v+v[itex]_{w}[/itex])/(v-v[itex]_{b}[/itex])]f[itex]_{b}[/itex], as mentioned above. And for the frequency heard by the bat, it's [(v+v[itex]_{b}[/itex])/(v-v[itex]_{w}[/itex])]f[itex]_{w}[/itex], where f[itex]_{w}[/itex] is the reflected, higher frequency.

    I'm so confused!
     
  9. Dec 8, 2011 #8

    PeterO

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    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Where did that frequency value of 11700 come from????
     
  10. Dec 8, 2011 #9
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Ah, my mistake! I'm working a problem with slightly different given frequencies (than the ones I've posted) so that I'd ensure I would figure out how to work it in another instance.

    That being said, ignore 11700. So, for this problem, it would be 22190=(the velocity part written above)x22000?
     
  11. Dec 8, 2011 #10

    PeterO

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    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Also, I suspect this can be done in one go.

    When you approach a mirror at 3 m/s, you image approaches you at 6 m/s [that is all with light]

    Acoustically I think this is similar.

    We have a bat approaching a wall at speed V. That means its image is approaching it at speed 2V.

    The bat, and the bats image are producing a frequency of 22000 Hz.
    The bat hears its own frequency, plus the frequency form the "approaching bat" with the doppler shift resulting in 190 beats per second.
     
  12. Dec 8, 2011 #11
    Re: The Doppler Effect -- frequency of a bat reflected on wall

    Ok, that much makes sense, although, I still do not at all see how to go about solving this.
     
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