Calculating the Frequency of a Tuning Fork Using Resonance Tube Method

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SUMMARY

The frequency of a tuning fork can be calculated using the resonance tube method, where the air column lengths are 48.5 cm and 68.7 cm at a temperature of 26˚C. The speed of sound in air at this temperature is approximately 346.6 m/s. The wavelength (λ) is determined to be half the difference between the two resonance lengths, calculated as λ = 0.404 m. Consequently, the frequency (f) of the tuning fork is found to be 857.4 Hz using the formula f = v / λ.

PREREQUISITES
  • Understanding of wave mechanics and sound propagation.
  • Familiarity with the resonance tube method for measuring sound frequencies.
  • Knowledge of the relationship between frequency, wavelength, and velocity (f = v / λ).
  • Basic principles of thermodynamics related to sound speed in air at varying temperatures.
NEXT STEPS
  • Study the effects of temperature on the speed of sound in air using the formula v = (332.5)(√(T/273)).
  • Explore the differences between open and closed pipe resonance and their impact on frequency calculations.
  • Learn about the practical applications of resonance tube experiments in acoustics.
  • Investigate advanced sound frequency measurement techniques, such as using digital oscilloscopes.
USEFUL FOR

Students in physics, acoustics researchers, educators teaching sound wave properties, and anyone interested in experimental methods for measuring sound frequencies.

Shackleford
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6. A tuning fork is heard to resonate over a resonance tube when the
air column is 48.5 cm long and again when it is 68.7 cm long. What is
the frequency of the tuning fork if the temperature is 26˚C?

I couldn't figure out how to work this. I thought it mattered if it were in a closed or open pipe, which is not mentioned.

v = 346.6 m/s
 
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I had a problem that was similar to this...if I recall correctly, lambda was .5(D2-D1). So lambda is half the distance between resonances.

>_> However, I am not an expert and could very easily be wrong.
 


Foxhound101 said:
I had a problem that was similar to this...if I recall correctly, lambda was .5(D2-D1). So lambda is half the distance between resonances.

>_> However, I am not an expert and could very easily be wrong.

Don't I have to know if it's open or closed?
 


lambda = 2 (l2 - l2) - .404 m

v = 346.4 ms^-1

f = 857.4 Hz
 


First you ll need to get the v-air which is given by: v = (332.5)(squareroot(T/273))
where T is the temperature in Kelvin. After that you know that v = f x lambda, and now i think: Ln+1 - Ln = 1/2 lambda.
 

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