What is the relationship between string tension and frequency on a ukulele?

AI Thread Summary
The discussion focuses on the relationship between string tension and frequency in a ukulele. When the string tension is increased, the frequency changes from 431 Hz to 444 Hz, leading to a calculation that indicates a 6% increase in force required. Participants confirm that the wavelength remains constant since the string length does not change and they are considering the fundamental frequency. The equation relating the wavelength to the string length is clarified, emphasizing that for the fundamental frequency, the harmonic number is fixed at one. Overall, the tension increase results in a proportional change in frequency while maintaining a consistent wavelength.
Drizzy
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Homework Statement


You have a ukulele and you change the strings tension so that it becomes tighter. First the frequency was 431 Hz and after we tightened the string the frequency. how much bigger does the force have t be to get that frequency?

Homework Equations



PA*V^2 = F (P = density, A = area)
V=f*lambda

The Attempt at a Solution


F2 = PA V2=PAf22 * lambda2
F1 = PA V2=PAf12 * lambda2
Then i divided F2 by F1 and I got:
f22/f12
which equals to 1,06. My answer is 6 precent bigger.

So my questions are: can I assume that lambda is going to remain the same? Is my solution correct?
 
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If lambda here represents the wavelength of the wave in the string, yes you can assume it remains the same. The string length did not change, and we are only interested in the fundamental.
You seem to have omitted the new frequency from the problem statement.
 
Sorry, the new frequency is 444Hz. Why can we assume that? That was my initial thought but then I thought wouldn't the wavelength differ if the string is tighter?
 
Drizzy said:
Sorry, the new frequency is 444Hz. Why can we assume that? That was my initial thought but then I thought wouldn't the wavelength differ if the string is tighter?
What is the equation relating the wavelength of the nth harmonic to the length of the string?
 
it is l=n*lambda/2
 
Drizzy said:
it is l=n*lambda/2
Right. So if we are only interested in the fundamental, that fixes n as 1. And the length l of the string does not change. So what does that tell you about lambda?
 

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