could anyone help me with physics? let''s say i have a violin string that is 40 cm long. (both ends are fixed). the string has a mass of 2.0 g. The open string (no finger applied) sounds an A note (440 Hz).. Now i want to play a C note (528 Hz). my question is: how far down the string do i place my finger? ----------------------------------------------------------------------- okay, i believe this has something to do with the new wavelength. I know that regardless of the frequencies, wave speed always remain the same. So my first intuition is to use the formula: freq. = n (v/(2L)) and solve for v (speed)... However, this does not seem to help I think i''m very close.. could anyone give me a hint as to how to go about solving this problem? thanks in advance!

Well once you know the frequency in Hz, you can find the wavelength since the speed of sound is constant anyhow. For example, you 528 Hz C note has a wavelength of 25.681818181818183 inches, 2.140151515151515 feet, or 0.6515151515151515 meters. Now I really don''t know what to do with the information now, since I don''t remember a lot of this from way back when in school, but that information will help you determine where to place your finger. Oh and I don''t remember the formula for calculating the frequency (i think its c = [lambda]v ).

This is the site I used to get the numbers, just in case you want to use it too: http://www.mcsquared.com/wavelength.htm

Dear Sir/Madam:

>For example, you 528 Hz C note has a wavelength of 25.681818181818183
>inches, 2.140151515151515 feet, or 0.6515151515151515 meters.

hehe..

I think you forgot one thing.. In this case, the medium is the
"string," not air. So I''m afraid your numbers dont help here.. ;P
(the speed is a property of string)

Still, thanks for your answer

I''m still trying to determine how the mass fit into the picture..
hmm.. this is really tricky..

ohh.. i''m still stumped...

actually, the formula is [wave speed] = [lambda] x [frequency]

i was thinking in line of resonance...

freq. = n (v/(2L)) --> n is an integral number..

time to give up..
good night!

PS: as you might have guessed, this is actually a homework problem.. ;P
it''s due in 7 hours and I''m just getting desperate..
sorry for the off-topic post.. ;P

Ha! got it!

Woooooooshhhhooooooooooooooo!!

this is the happiest day of my life!!!!!!!!!

time to hit the sack..

byeeeee!!!!!!

Just out of interest, what is the solution?

If I remember correctly, the transverse wave speed is constant in a string as long as the tension and the mass per unit length are constant. So, using the information for an A, at 440Hz, and that the string is 40cm long, the speed of the wave is:

c = 2LF = 2 * 40cm * 440/s = 35200cm/s

then, at F = 528Hz:

L = c/2F = 35200cm/s / (2*528/s ) = 33.33cm

So, you would want to put your finger about 7cm down the string.

>Just out of interest, what is the solution?

My first intuition was correct.
this is a problem about resonance..

the trick is using this formula:

freq. = n (v/(2L)) --> n is an integral number..
(this formula relates frequency to wave velocity and
the total string length)

to find the specific harmonic numbers (n) at the given frequencies:

440HZ (note A) and 580 HZ (note C)

Also, the key to solve this problem is in recognizing that:

1) wave velocity is a property of string; thus, the wave velocity
is independent of frequencies..
2) Note C is two steps up from Note A. If note A is at a specific
harmonic n, note C would be n+2.

now we wanna solve this equation for v (velocity), plug the numbers
in two separate equations, and set the two equations equal to each
other:

[velocity] = (2L * freq.) / n

[(2 * .4 [meter] * 440 [HZ]) / n] = [(2 * .4 * 580) / n+2]

solve for n...

I think i got n = 10 (something like that)

then, the specific harmonic at 580HZ is (n+2) = 12.

Finally, we use this information to find lambda (the wavelength)
using this formula:

lambda = 2L / n

in this case, n = 12...

Vola.. you got lambda..

then by subtracting the number from the total string length,
you can determine where to place your finger

Antknei:

i believe the answer is somewhere close to 30 cm

>c = 2LF = 2 * 40cm * 440/s = 35200cm/s

this wouldnt work, cuz you cant find wave speed directly
from frequency

the definition of frequency is number of cycles per second,
which is inverse of period..
frequency is also related to wave speed by this equation:

lambda (wavelength) x freq. = wave speed

There's an additional piece of information we need before
we can find wave speed from frequency. (the wavelength!)

is 40 cm the wavelength? No, it's the total length of
string.. (wavelength is the information we're trying
to find!)

So, i believe the only way to obtain the answer is by
following the procedure i described in my previous post

take care...

Edited by - Sandwichlover on January 17, 2001 6:45:20 PM