A question about a vibrating string

In summary, the conversation discusses how vibrato is produced on a violin by sliding the finger back and forth along the vibrating string. The G-string on a particular violin measures 30cm and has a fundamental frequency of 197Hz. The conversation then goes on to calculate the distance a finger should be placed from the end of the string in order to play the note A(440Hz) and the resulting range of frequencies when executing vibrato by moving the finger 0.50 cm to either side. The correct answer for the first part is 0.166m, and for the second part, the range of frequencies is 424-457Hz.
  • #1
jianghan
25
0

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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  • #2
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.
 
  • #3
I did it slightly different and got an answer of 0.134m too.
 
  • #4
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
 
  • #5
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
 

Related to A question about a vibrating string

1) What causes a string to vibrate?

A string vibrates due to the energy from a source, such as plucking or striking the string, causing it to move back and forth at a certain frequency.

2) How does the thickness of the string affect its vibration?

The thickness of a string affects its vibration by determining its tension and density, which in turn affects the frequency and amplitude of the vibrations produced.

3) What is the relationship between string length and pitch?

The shorter the string, the higher the pitch will be, and vice versa. This is because a shorter string has a higher frequency of vibrations compared to a longer string.

4) Can different materials produce different vibrations in a string?

Yes, different materials can produce different vibrations in a string. The type of material affects the density, tension, and stiffness of the string, which all contribute to the resulting vibrations.

5) How do harmonics play a role in the vibration of a string?

Harmonics are the natural frequencies of a string that can be produced when the string is divided into equal sections. These harmonics contribute to the overall vibration pattern of the string and can create different sounds and tones.

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