If we had a pipe which could play two successive harmonics 240Hz and 280Hz, how long would it have to be? Open ended or closed ended?
I know that $f_n=nf_1$ for open end and $f_n=nf_1 $for n is odd and >=1. So how do I decide weather it's open or closed?
What I tried was that I pretended that it was open and saw that the difference between the frequency was 40. So with that I can conclude the fundamental frequency must be 40 Hz if it was open. If we do assume it is opened we can say for the second harmonic that $v=fλ$ which is$340=λ80$ then $λ= 4.25= length$ However how do I know if it is actually open or closed?
Then I tried the something for closed ended. From what I know the 2nd harmonic must be 3x the frequency of the fundamental and that 3rd harmonic is 5x the fundamental and 4th harmonic is 7x …. However that means the difference in frequency between each harmonic is different.
The last piece of information I have is that regardless weather the pipe is open or closed the frequency of each harmonic is a multiple of the fundamental frequency.
Please correct me if I have any misconceptions regarding waves and please explain how I can solve this problem.
Best Answer
As you say, for both pipes (open at both ends and closed at one end) the formula $f_n=nf_1$ applies. For open pipes $n$ is any integer, and the separation between successive harmonics is $f_1$. For closed pipes $n$ is an odd integer, and the separation between successive harmonics is $2f_1$.
First you need to find $f_1$ in both cases. Then decide whether the two frequencies given are successive integer multiples or successive odd-integer multiples of $f_1$. You will then be able to decide whether the pipe is open or closed. Finally, knowing $f_1$ and whether the pipe is open or closed and the speed of sound in air, you can work out the length of the pipe.