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

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SUMMARY

The discussion centers on the relationship between string tension and frequency in a ukulele, specifically analyzing how tightening the strings affects frequency. Initially, the frequency was 431 Hz, which increased to 444 Hz after tightening. The calculations indicate that the force required to achieve this frequency increase is approximately 6% greater. The wavelength remains constant as the string length does not change, allowing for the assumption that the fundamental frequency's wavelength is fixed.

PREREQUISITES
  • Understanding of wave mechanics, specifically frequency and wavelength relationships.
  • Familiarity with the physics of string instruments and tension effects.
  • Knowledge of basic equations relating pressure, area, and force (PA*V^2 = F).
  • Ability to manipulate equations involving harmonic frequencies and wavelengths.
NEXT STEPS
  • Research the physics of string tension and its impact on frequency in string instruments.
  • Explore the relationship between harmonic frequencies and string length in greater detail.
  • Learn about the effects of material properties on string vibration and tension.
  • Investigate advanced wave equations and their applications in musical acoustics.
USEFUL FOR

Musicians, physics students, and string instrument builders interested in understanding the acoustic properties of ukuleles and the impact of string tension on sound production.

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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