- #1
Strbarrytree
- 2
- 0
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
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