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Waves of sound from singer

  • Thread starter amit25
  • Start date
30
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1. Homework Statement
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. Homework 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.
 

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?)
 
30
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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
 

tiny-tim

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if you want us to check your work, you'll have to show us your calculations. :wink:
 
30
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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
 

tiny-tim

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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
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?
 
30
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oh okay im just trying to do it the way my teacher did
but the ratio is 132cm/65.9cm=2
 

tiny-tim

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30
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233Hz/294Hz=0.80
which is less than the wavelength ratio
 
30
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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
 

tiny-tim

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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
where do 927 and 343 come from? :confused: :confused:
 
30
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927 is the speed of helium and 343 is the speed of sound
 

tiny-tim

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what were the 330 and 965 you used before? :confused:
 
30
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those are the same values i just researched to get more accurate values
 

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?
 
30
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so 2.7 x 262hz=707hz so thats F5 note ?
 

tiny-tim

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so 2.7 x 262hz=707hz so thats F5 note ?
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:
 
30
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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:

tiny-tim

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from here not sure which octave
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?
 
30
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lol well i have a sheet we got with all the notes and frequency so i dont think were expected to memorize them, anyways thanks i feel like this is going no where ill just ask my professor
 

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