Find lowest resonant frequence on this string

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SUMMARY

The lowest resonant frequency of a string stretched between fixed supports of 75 cm is 315 Hz, while the next resonant frequency is 420 Hz. The absence of intermediate resonant frequencies indicates that these correspond to the fundamental frequency and the first harmonic. To determine the wave speed, the formula f_n = nv/2L can be utilized, where n represents the harmonic number, v is the wave speed, and L is the length of the string.

PREREQUISITES
  • Understanding of harmonic frequencies in vibrating strings
  • Familiarity with wave speed calculations
  • Knowledge of fundamental frequency concepts
  • Ability to solve systems of equations
NEXT STEPS
  • Study the relationship between harmonic frequencies and wave speed using the formula f_n = nv/2L
  • Explore the concept of standing waves in strings
  • Learn about the effects of tension and mass per unit length on resonant frequencies
  • Investigate practical applications of resonant frequencies in musical instruments
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Physics students, music educators, and engineers interested in acoustics and wave mechanics will benefit from this discussion.

Saladsamurai
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A string that is stretched between fixed supports separated by 75 cm has resonant frequencies of 420 and 315 Hz, with no resonant intermediate frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Okay: So I know that since there are no intermediate resonant frequencies, then the two givens are of nth harmonic and nth+1 harmonic.

I also know that [tex]f_n=nf_1[/tex] and [tex]f_{n+1}=(n+1)f_1[/tex]

so I need to find the fundamental frequency right? I am assuming this will involve 2 equations in 2 unknowns...
 
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Should I use [tex]f_n=\frac{nv}{2L}[/tex] in a system of equations to solve for v and then plug that into find the 1st harmonic?...how bout I try it instead of asking?!
 

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