Calculating the Frequency of a Tuning Fork Using Resonance Tube Method

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Homework Help Overview

The problem involves calculating the frequency of a tuning fork using the resonance tube method, with specific air column lengths provided and temperature noted. The context is within the subject area of wave mechanics and acoustics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the lengths of the air column and the wavelength, with some suggesting that the wavelength is related to the difference in resonance lengths. Others express uncertainty about whether the pipe is open or closed, which may affect the calculations.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have provided formulas and calculations, while others are questioning the assumptions regarding the type of pipe and the implications for the frequency calculation.

Contextual Notes

There is mention of specific temperature conditions and the speed of sound in air, which may influence the calculations. Participants are also considering the impact of whether the resonance tube is open or closed, which has not been clarified in the original problem statement.

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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