Calculating the Speed of Waves in a Vibrating String and Resonant Pipe

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SUMMARY

The discussion centers on calculating the speed of transverse waves on a vibrating string and its interaction with a resonant pipe. A 0.13-meter string vibrating in its fourth harmonic excites a 0.88-meter pipe in its second overtone resonance. The correct formula used is F=nv/(2L), leading to the conclusion that the speed of waves on the string is 67.6 m/s. However, discrepancies in calculations indicate a misunderstanding of the relationship between the frequencies of the string and pipe waves.

PREREQUISITES
  • Understanding of harmonic frequencies in vibrating strings
  • Knowledge of wave speed calculations using the formula F=nv/(2L)
  • Familiarity with the concept of resonant frequencies in open pipes
  • Basic grasp of sound speed in air, specifically 345 m/s
NEXT STEPS
  • Review the principles of harmonic motion in strings and pipes
  • Learn about the relationship between frequency and wave speed in different mediums
  • Study the derivation and application of the formula F=nv/(2L) in various contexts
  • Explore the concept of overtone resonances in musical acoustics
USEFUL FOR

Students in physics or engineering, educators teaching wave mechanics, and anyone interested in the principles of sound and vibration in strings and pipes.

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Homework Statement


A 0.13-m string, fixed at both ends and vibrating in its n = 4 harmonic, excites a pipe that is 0.88 m long and open at both ends, into its second overtone resonance. What is the speed of transverse waves on the string? The speed of sound in air is 345 m/s.

Homework Equations


F=nv/(2L)

The Attempt at a Solution


I wasn't present when my class went over this, so keep in mind that this is just my speculation. Anyway, as the string excites the pipe, the frequencies of both waves should be the same.
f1=f2
I don't know the velocity of the first wave, but the velocity of the second wave should be the speed of sound. Now, plugging in numbers I know, I get
4(v)/(2*.13m)=3(345 m/s)/(2*.88m)
v=67.6 m/s

The online quiz I'm doing is telling me this is wrong, so what's wrong with my logic?
 
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JSmithDawg said:
4(v)/(2*.13m)=3(345 m/s)/(2*.88m)
v=67.6 m/s

I don't get 67.6 m/s when solving your equation for v.
 

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