Calculating the Frequency of a Tuning Fork Using Resonance Tube Method

[Shackleford]

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

 

[Foxhound101]

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.

 

[Shackleford]

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?

 

[Shackleford]

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

v = 346.4 ms^-1

f = 857.4 Hz

 

[abelanger]

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.