A question about a vibrating string

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To play the note A (440Hz) on a 30cm G-string, the violinist should place their finger 0.166m from the end of the string. The fundamental frequency is altered by changing the vibrating length of the string, which is calculated as 0.134m. The speed of the wave remains constant at 118.2m/s, leading to a wavelength of 0.269m. For the vibrato effect, moving the finger 0.50 cm to either side results in a frequency range of 424-457Hz. The discussion clarifies the relationship between string length, frequency, and vibrato execution.
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Homework Statement



"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?

Homework Equations


f=nV/2L=nf1

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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jianghan said:
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.
 
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
 
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