Harmonics on a Stretched Spring

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  • #1
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Homework Statement


A. A string with a mass density μ = 4.10×10-3 kg/m is under a tension of F = 327 N and is fixed at both ends. One of its resonance frequencies is 742.0 Hz. The next higher resonance frequency is 1113.0 Hz. What is the fundamental frequency of this string?

B. Which harmonic does the resonance frequency at 742.0 Hz correspond to? (i.e. what is n at this frequency?)


C. What is the length of the string?


Homework Equations





The Attempt at a Solution


So for A. I wanted to use the equation...
FF = 1/2L * sqrt(T/μ)
FF - fundamental frequency
Where...
μ = 4.10×10-3 kg/m
T = 327N
but I don't know what L is?
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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The standing waves in a string forms with integral multiples of half of the wavelength of the generated wave in the string

[tex]L = n\frac{\lambda}{2}[/tex]

and for the speed in the wave we can substitute

[tex]\lambda f = \alpha[/tex]

where

[tex]\alpha = \sqrt{\frac{T}{\mu}}[/tex]

giving

[tex] \lambda = \frac{1}{f} \alpha[/tex]

lets say that the one standing wave forms such that

[tex] L = \frac{n\lambda _1}{2}[/tex]

and the next harmonic such that

[tex] L = \frac{(n + 1) \lambda _2}{2}[/tex]

equating gives

[tex]\frac{n\lambda _1}{2} = \frac{(n + 1) \lambda _2}{2}[/tex]

therefore

[tex]n\lambda _1 = (n + 1) \lambda _2[/tex]

substituting now for [tex]\lambda[/tex] gives

[tex] \frac{n \alpha}{f_1} = \frac{(n+1) \alpha}{f_2} [/tex]

.......
 
Last edited:
  • #3
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Wow...okay so is that all just for the first part of the question??
 
  • #4
602
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Ok I got the answer for A and B (A was 371Hz and B was 2)....but I can not figure out how to determine the length of the string. I thought the equation to use was...
L = V/2f
but I don't have the velocity...
Can someone please help me with this.
Thank you
 

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