What is the length of the vibrating section of the violin string?

  • Thread starter Thread starter Deviatorz
  • Start date Start date
  • Tags Tags
    Standing wave Wave
Click For Summary

Homework Help Overview

The discussion revolves around determining the length of the vibrating section of a violin string based on given parameters such as wavelength, speed of sound, linear density, and tension. The subject area includes concepts from wave mechanics and harmonic motion in stringed instruments.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between tension, linear density, and wave speed. Questions arise regarding the definition of the "vibrating section" and its dependence on the harmonic being played. Some participants express confusion about the calculations and the necessary conversions for linear density.

Discussion Status

The discussion is ongoing, with participants sharing their interpretations and clarifications about the vibrating section and harmonics. Some guidance has been offered regarding the relationship between string length and harmonics, but no consensus has been reached on the specific calculations needed.

Contextual Notes

There is a noted confusion regarding the units of linear density, as participants mention the need to convert from grams per meter to kilograms per meter. The implications of different harmonics on the vibrating length are also under consideration.

Deviatorz
Messages
6
Reaction score
0

Homework Statement


A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightly warm, so the speed of sound is 344 m/s.
If the string's linear density is 0.620 g/m and the tension is 160 N, how long is the vibrating section of the violin string?

T = 160N
Linear Density = 0.62g/m
Wavelength of violin string = 0.391m
Speed of sound in warm room = 344m/s
m = 1?

Homework Equations



v = \sqrt{\frac{T}{\mu}}

v = \lambda\timesf

The Attempt at a Solution



I all know is to find the velocity of the string first which is the square root of 250 or about 15.811m/s. I am unsure of what to do next. Also, what is the "vibrating section of the violin string"? I am unsure of what that is. Thanks in advanced.
 
Physics news on Phys.org
Deviatorz said:

Homework Statement


A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightly warm, so the speed of sound is 344 m/s.
If the string's linear density is 0.620 g/m and the tension is 160 N, how long is the vibrating section of the violin string?

T = 160N
Linear Density = 0.62g/m
Wavelength of violin string = 0.391m
Speed of sound in warm room = 344m/s
m = 1?

Homework Equations



v = \sqrt{\frac{T}{\mu}}

v = \lambda\timesf

The Attempt at a Solution



I all know is to find the velocity of the string first which is the square root of 250 or about 15.811m/s. I am unsure of what to do next. Also, what is the "vibrating section of the violin string"? I am unsure of what that is. Thanks in advanced.



The vibrating section of the string is that of the standing wave, i.e. the bit better the bridges of the violin. This length is dependent on the note played. To solve this problem you need to figure out which harmonic the string is vibrating at.
 
Erwin Kreyszig said:
The vibrating section of the string is that of the standing wave, i.e. the bit better the bridges of the violin. This length is dependent on the note played. To solve this problem you need to figure out which harmonic the string is vibrating at.

I still don't quite get what the "vibrating section" of the violin string is. Are you saying that it is the length of the string; which makes no sense. I think the harmonic for stringed instruments is one or m = 1. Somewhere in between I have to use this formula:

f1 = \frac{1}{2L}\times\sqrt{\frac{T}{\mu}}.
 
Deviatorz said:
I still don't quite get what the "vibrating section" of the violin string is. Are you saying that it is the length of the string; which makes no sense. I think the harmonic for stringed instruments is one or m = 1. Somewhere in between I have to use this formula:

f1 = \frac{1}{2L}\times\sqrt{\frac{T}{\mu}}.

With stringed instruments you have as many harmonics as there are notes, effectively that is the physics behind it. A violinist can place his finger very lightly on a string so that it divides the string into half. He will hear a harmonic (the note an octave higher than the open string). By placing his fingers in other places he can get more harmonics, e.g by touching the string a quarter of the way down he gets the next harmonic.
 
Sorry for the late reply, the way I approached the question was correct. My answer was incorrect because the linear mass density in the question was in g/m while I needed to convert to kg/m. This question took me hours to figure out, I was sure the way I was doing was correct.

For future references, the answer to this question is 28.9cm.
 

Similar threads

Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
10K
  • · Replies 26 ·
Replies
26
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
4
Views
13K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K