Using frequency to calculate resonances

[jimisreincarn]

1. If the length of the tube used in our experiment is 4 m, how many resonances would you observe when a tuning fork of frequency 256HZ is used?



2. v=f\lambda ; L = (1/4)(2n+1)\lambda



3. 340m/s = 256Hz\lambda \lambda = 1.328125m
4 = (1/4)(2n+1)(1.328125m)
12.05 = 2n+1
5.5 = n

 

[AtticusFinch]

We need to know what kind of tube you used in lab (open, one end closed, etc.)

 

[jimisreincarn]

one end of the tube is open. The other end has water at the end of it.

 

[AtticusFinch]

one end of the tube is open. The other end has water at the end of it.

Ok and you probably raised the water level while ringing the tuning fork and marked where you heard resonances, right?

So this is a closed end tube. You are correct in using the equation L = \frac{2n+1}{4}\lambda

However, you are not correct in leaving L at 4 because you raised and lowered the water level. What you should do is evaluate the inequality 4 \leq \frac{2n+1}{4}\lambda
and solve for n. Of course you are only allowed integer values of n.

 

[jimisreincarn]

okay, that makes sense. so if my evaluation of the wavelength is correct, i would substitute it into the inequality and I should get n < 5? so at most there are 5 resonance structures.

 

[AtticusFinch]

okay, that makes sense. so if my evaluation of the wavelength is correct, i would substitute it into the inequality and I should get n < 5? so at most there are 5 resonance structures.

Yep, that is what I get as well.