Question

One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 hz, and the next higher harmonic has a frequency of 576 hz. what is the fundamental frequency of the air column in this tube?

Answer 1

The general formula for the frequency of the nth-harmonic of the column of air in the tube is given by
$f_n = n f_1$
where f1 is the fundamental frequency.

In our problem, we have two harmonics, one of order n and the other one of order (n+1) (because it is the next higher harmonic), so their frequencies are
$f_n = n f_1$
$f_{n+1} = (n+1) f_1$
so their  difference is
$f_{n+1} - f_n = (n+1) f_1 - n f_1 = (n+1-n) f_1 = f_1$

So, the difference between the frequencies of the two harmonics is just the fundamental frequency of the column of air in the tube, which is:
$f_1 = 576 Hz - 448 Hz = 128 Hz$