Tripling fundemental frequency

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SUMMARY

The fundamental frequency of a bass violin string at 1045 Hz occurs when the string is 0.900 m long. To achieve a frequency three times greater, the vibrating length of the string must be reduced to one-third, resulting in a length of 0.300 m from the lower fixed end. The discussion highlights a disagreement regarding the relevance of the string's length versus the speed of sound in air, with the consensus being that the string's resonant frequency is independent of air properties. The correct approach involves analyzing the string as a transmission line with a fixed velocity factor.

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  • Understanding of fundamental frequency and harmonics in string instruments
  • Knowledge of wave propagation and standing waves
  • Familiarity with the concepts of tension and loading in strings
  • Basic principles of acoustic physics, including the speed of sound
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  • Research the relationship between string length and frequency in string instruments
  • Study wave propagation in transmission lines and its application to musical instruments
  • Learn about the effects of tension and mass per unit length on string vibration
  • Explore advanced concepts in acoustics, such as harmonics and resonance
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Strbarrytree
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So the question goes like this- The fundamental frequency of a bass violin string is 1045 Hz and occurs when the string is 0.900 m long. How far from the lower fixed end of the bass violin should you place your fingers to allow the string to vibrate at a fundamental frequency 3 times as great?

I say you take (2*.9)*1045 to get 1881 m/s because of v=f*w, then you take (1881/(1045*3))/2 to get .3 because of f=v/2l, and .3 would be the answer which would make sense since .3=(1/3).9

My teacher who is a smart man says you would disregard the information you got about the length of the string (the 0.900 m) and just take (340(speed of sound in air)/(3*1045))/2 or something like that he got like .84 or .084 which doesn't add up to the equation I just said, but that's what he said to do because he said it has to travel through air so we can hear it.

I think that that is irrelevant because it is not asking for how we perceive the frequency but rather what the string is actually doing. So I just kinda want an answer to know whose right so if this is on my text I know what to do. Thank you
 
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The resonant frequency of a string is independent of the air surrounding the string.

The weight or loading of a string per unit length will be constant, we assume the tension in the string also remains constant. You can analyse that string as a transmission line with a fixed velocity factor supporting a standing wave.

To get three times the frequency you must reduce the transit time to 1/3 of the original. That says the excited section of the string needs to be only 1/3 as long.

An alternative might be to bow it at 1/6 of the length while killing the fundamental and second harmonics with a finger at the 1/3 point. Then there would be three cycles of third harmonic wave propagating on the full length of the line or string.
 

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