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Waves of sound

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Singer sings middle C. She then gulps helium and tries to sing the same note i.e her vocal chords remain the same. What note will the listener hear?





    2. Relevant equations
    f=velocity of air/lambda middle C



    3. The attempt at a solution
    f=velocity of air/lambda middle C
    gives lambda=330m/s/262Hz=1.2595

    After this point im lost i think you have to use freq of helium=velocity of helium/lambda of middle C
    but what note is that?
    is it suppose to be C 4 octaves higher or the note G.
     
  2. jcsd
  3. Jun 23, 2012 #2

    tiny-tim

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    hi amit25! :smile:

    you don't need to find the frequencies, they cancel out!

    the frequencies are the same, but the velocities and wavelengths are different, so what is the equation relating them? :wink:

    (and what is the ratio of wavelengths of middle C and the next C?)
     
  4. Jun 23, 2012 #3
    the steps i did before is what my teacher did, he just solved it halfway and said thats the equation we use, so im a bit confused because i calculated it to be 766Hz which is around the note G and then after some research i found that the note is still C but i guess a few octaves higher? not sure...really confused
     
  5. Jun 23, 2012 #4

    tiny-tim

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    if you want us to check your work, you'll have to show us your calculations. :wink:
     
  6. Jun 23, 2012 #5
    okay this is what i did

    f=velocity of air/lambda middle C

    262Hz=330m/s / lambda middle C

    which gives lamba=1.2595 m

    frequency of helium is f=velocity of helium/lambda

    f=965m/s/1.2595m
    f=766Hz
     
  7. Jun 23, 2012 #6

    tiny-tim

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    ok so far (but it would be a lot quicker not to find either of the wavelengths)

    now what is the ratio of wavelengths of middle C and the next C?
     
  8. Jun 23, 2012 #7
    oh okay im just trying to do it the way my teacher did
    but the ratio is 132cm/65.9cm=2
     
  9. Jun 23, 2012 #8

    tiny-tim

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    now compare that with the ratio of the two wavelengths in the question :smile:
     
  10. Jun 23, 2012 #9
    233Hz/294Hz=0.80
    which is less than the wavelength ratio
     
  11. Jun 23, 2012 #10

    tiny-tim

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    where do those figures come from? :confused:
     
  12. Jun 23, 2012 #11
    Wavelength in helium λ = c / f = 927 / 262 = 3.538 m.
    Wavelength in air λ = c / f = 343 / 262 = 1.309 m.

    Ratio = 3.538 / 1.309 = 2.7
     
  13. Jun 23, 2012 #12

    tiny-tim

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    where do 927 and 343 come from? :confused: :confused:
     
  14. Jun 23, 2012 #13
    927 is the speed of helium and 343 is the speed of sound
     
  15. Jun 23, 2012 #14

    tiny-tim

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    what were the 330 and 965 you used before? :confused:
     
  16. Jun 23, 2012 #15
    those are the same values i just researched to get more accurate values
     
  17. Jun 23, 2012 #16

    tiny-tim

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    ok, if the frequency ratio is 2.7, and the lower note is middle C, then how do we work out the higher note?
     
  18. Jun 23, 2012 #17
    so 2.7 x 262hz=707hz so thats F5 note ?
     
  19. Jun 23, 2012 #18

    tiny-tim

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    what's F5 ? :confused: do you mean F# ? and which octave ?

    and how did you work it out?

    or did you just look it up in a table? :redface:
     
  20. Jun 23, 2012 #19
    from here not sure which octave anyways i think its wrong the answer should still be C
    http://www.sengpielaudio.com/calculator-… [Broken]
    Scroll down to "Frequency to Musical Note Converter".
     
    Last edited by a moderator: May 6, 2017
  21. Jun 23, 2012 #20

    tiny-tim

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    you won't be able to take that website into the exam with you :redface:

    (btw, that link isn't working)

    you need to know how to calculate it yourself

    if the ratio was 2 (instead of 2.7), what would the note be?

    if the ratio was 4, what would the note be?
     
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