# Waves of sound from singer

#### amit25

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.

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#### tiny-tim

Homework Helper
hi amit25! 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? (and what is the ratio of wavelengths of middle C and the next C?)

#### amit25

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

Homework Helper
if you want us to check your work, you'll have to show us your calculations. #### amit25

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

Homework Helper
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?

#### amit25

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

#### tiny-tim

Homework Helper
… the ratio is 132cm/65.9cm=2
now compare that with the ratio of the two wavelengths in the question #### amit25

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

#### tiny-tim

Homework Helper
233Hz/294Hz
where do those figures come from? #### amit25

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

Homework Helper
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?  #### amit25

927 is the speed of helium and 343 is the speed of sound

#### tiny-tim

Homework Helper
what were the 330 and 965 you used before? #### amit25

those are the same values i just researched to get more accurate values

#### tiny-tim

Homework Helper
ok, if the frequency ratio is 2.7, and the lower note is middle C, then how do we work out the higher note?

#### amit25

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

#### tiny-tim

Homework Helper
so 2.7 x 262hz=707hz so thats F5 note ?
what's F5 ? do you mean F# ? and which octave ?

and how did you work it out?

or did you just look it up in a table? #### amit25

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".

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#### tiny-tim

Homework Helper
from here not sure which octave
you won't be able to take that website into the exam with you 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?

#### amit25

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