## A question about a vibrating string

1. The problem statement, all variables and given/known data

"Vibrato" in a violin is produced by sliding the finger back and forth along the vibrating string. The G-string on a particular violin measures 30cm between the bridge and its far end and is clamped rigidly at both points. its fundamental frequency is 197Hz
(a) how far from the end should the violinist place a finger so that the G-string plays the note A(440Hz)?
(b)If the violinist executes vibrato by moving the finger 0.50 cm to either side of the position in part(a),what is the resulting range of frequencies?

2. Relevant equations
f=nV/2L=nf1

3. The attempt at a solution
i assume speed of wave would remain the same so v=f1x(2L)=118.2m/s
λ=118.2/440=0.269m
as the rigid end of the string is node,so the finger should be put at next node which is 0.5λ away,that is 0.269/2=0.134m, but the answer is 0.166m. also i think n should be whole numbers but in this case i couldn't get n to be a whole number.

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 Quote by jianghan [b]i assume speed of wave would remain the same so v=f1x(2L)=118.2m/s λ=118.2/440=0.269m as the rigid end of the string is node,so the finger should be put at next node which is 0.5λ away,that is 0.269/2=0.134m, but the answer is 0.166m. also i think n should be whole numbers but in this case i couldn't get n to be a whole number.
You have determined the speed of propagation correctly. However, then you got confused by thinking about harmonics. While in this case, by placing a finger on the string, the violinist changes the length of the string and, consequently, the fundamental frequency.

 I did it slightly different and got an answer of 0.134m too.

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## A question about a vibrating string

0.134 m is the vibrating length of the string, between the bridge and the finger. So the finger has to be placed at (0.3-0.134)=0.166 m distance from the end of the string.

ehild

 ok,thank you all for your help! i think i can do both parts now. the second part ans is 424-457Hz which is relatively easy