# Closed ended tube resonance problem

1. Apr 17, 2013

### AbsoluteZer0

1. The problem statement, all variables and given/known data

In class a 3 m long aluminum bar was made to resonate by clamping it at a node in the centre of the bar. The frequency heard was 1200Hz.

2. Relevant equations

$L = \frac{n}{4}\lambda$

$v = f\lambda$

3. The attempt at a solution

Knowing that L is 3, we can substitute this into the formula and solve accordingly in order to find wavelength:

$3 = \frac{1}{4}\lambda$
$(4)(3) = \lambda$

Which gives us 12m. We can then use the wave equation, substituting 12 and 1200

$v = f\lambda$
$v =(1200)(12)$

to arrive at $14400 m s^-1$

However, I am lead to believe that I am possibly wrong due to the relatively high value of velocity that I found. Am I wrong? If so, where is my mistake?

Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 17, 2013

### TSny

What must be at each end of the rod - a node or an antinode? Rethink the expression for $L$ in terms of $\lambda$.

3. Apr 17, 2013

### AbsoluteZer0

I believe an antinode at each end, making it an open ended tube.

$L = \frac{n}{2}\lambda$

L is therefore 6 m, and velocity is therefore 7200 m s-1

Is this correct?

Thanks,

4. Apr 17, 2013

### TSny

Yes, that looks right to me. (According to a quick check on the internet, 7200 m/s is somewhat high for aluminum. But that's what your data gives.)

5. Apr 17, 2013

Thanks