Wavelength of the 2nd Overtone of a string.

[MissKaylaPaige]

Homework Statement

Find the wavelength of the 2nd overtone of a string 4.5 m long between points with a transverse speed of 48 m/s.

Homework Equations

The Attempt at a Solution

Since the fundamental frequency is 9, shouldn't the 2nd overtone be 2/3 of that? I'm completely baffled!

 

[MissKaylaPaige]

Please??

 

[Dick]

Ok, I'll try. Shouldn't the second overtone be three times the fundamental frequency? Furthermore, why should the fundamental frequency be '9'? Use units, please! Isn't the fundamental wavelength 9m? I think you are confusing wavelength and frequency.

 

[Mindscrape]

I don't remember what an overtone is because it is worthless, all you care about are eigenfrequencies and harmonics, but I think she is right.

The first overtone, unless I am getting it confused with something else, would be if you were to divide the length of the string into half. The second would be if you were to divide it into 1/3, the fourth of 1/4, and then you want the node before the end. So for the second overtone 1-1/3 = 2/3. Of course then you would have to multiply that factor by the total wavelength.

 

[MissKaylaPaige]

Yes, I meant the fundamental WAVELENGTH is 9m. The fundamental FREQUENCY is 5.3 Hz. But no, 3 times the fundamental frequency is not giving me the correct answer for the wavelength of the 2nd overtone.

 

[Dick]

If the fundamental wavelength is 9m, then the second overtone is wavelength 3m. That has 3 times the frequency of the fundamental. Now you've got Mindscrape confused. The fundamental frequency isn't 5.3Hz, it's 16/3 Hz which is 5.33333333...Hz. I think you are making a rounding error.

 

[Bob Dole]

Dick is correct. Wave speed = frequency * wavelength

48.0 m/s = freq * 9 m
freq= 48/9 = 5.33 Hz

I know this is a post 4 years later and you either figured this problem out or shot your professors dog. Either way, I am taking physics 3 now and i am looking up overtone as well :) and I am planning on shooting his dog :P

EDIT: I found out that MINDSCRAPER is correct, overtone is useless.

Overtone is related to harmonic by being 1 behind (aka. 2nd harmonic --> 1st overtone, 3rd harmonic --> 2nd overtone)
Therefore, 2nd overtone frequency = 3 * fundamental frequency
2nd overtone frequency = 3 * 9 m = 27 m
2nd overtone wavelength is found working backwards with the equation v = lambda * frequency