# Open and closed pipe

1. Dec 12, 2014

### diracdelta

1. The problem statement, all variables and given/known data
Open and closed pipe, give 5 beats per second.
Open pipe is 30 cm long and gives a tone of frequency f0.
Speed of sound is 330 m/s.
How long do we need to extend closed pipe so both pipes give equal frequencies?

2. Relevant equations
Open pipe, fn=n*f1, f1=v/(2L)
Closed pipe; fn= f*f1, f1 = v/(4L)
fbeat = f2 - f1
3. The attempt at a solution
For open pipe, I assume that f0 is f1, or basic frequency.
f=v/2L=330 ms-1 / 0,6 m = 550 Hz

fbeat = f2 - f1
5 = f2 - 550 Hz
f2 = 555 Hz.

Now we need to find the lenght of open pipe.
4L = v/f2
l= v/4f2= 0,15 m

Since i have the lenght of closed pipe, motive is to adjust it so we have frequency of 550 Hz.
4L = v/f2,
4*L = 330 ms-1/ 550 Hz
4*L = 0,6 m
L = 0,15

What did I do wrong?

2. Dec 12, 2014

### Staff: Mentor

I think you mean closed pipe? Better check that calculation, too, using 555Hz.

3. Dec 12, 2014

### ehild

You ment closed pipe, didn't you?

l= v/(4f2)=330/(4*555)=0.1486 m
You rounded off too early, too much.

You rounded off too much.

4. Dec 13, 2014

### diracdelta

Yes. lapsus linguae :D

:(
I did it again, i got same result like yours.

Thanks for help guys.
You are the best :)