Question

One of the harmonic frequencies of tube A with two open ends is 836 Hz. The next-highest harmonic frequency is 912 Hz. (a) What harmonic frequency is next highest after the harmonic frequency 228 Hz? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube B with only one open end is 2940 Hz. The next-highest harmonic frequency is 3220 Hz. (c) What harmonic frequency is next highest after the harmonic frequency 3780 Hz? (d) What is the number of this next-highest harmonic?

Answer 1

With Tube A having two open ends then the frequency will have an anti-node at both open ends.

This works for all tubes with two open ends, if you have two harmonic frequencies in succession then to get the first harmonic frequency just subtract the two. All harmonic frequencies will be multiples of the first harmonic frequency.
First harmonic frequency:

912 Hz - 836 Hz = 76 Hz

Great, we have the first harmonic (65 Hz).

So to answer part a)

228 Hz + the 1st harmonic (76 Hz) = the next harmonic (304 Hz)

The number of the next harmonic is pretty simple.

(b) The harmonic frequency / the 1st harmonic frequency = the Harmonic number
So:
304 / 76 = 4


For the tube with one closed end:
- only odd harmonic frequencies are possible (where the harmonic number is 1, 3, 5, 7,9, ...)
- there is a node at the closed end and an anti-node at the open end.

So the two harmonic frequencies that are given are actually 2 harmonic frequencies apart, since even number harmonic frequencies are not possibly in a tube with one open end.

So therefore, (one harmonic frequency - the harmonic frequency before it)/2 = the 1st harmonic frequency.

First harmonic frequency = (3220 - 2940)/2 = 140 Hz

Part C, we actually need to add the first harmonic frequency twice in order to skip the even harmonic frequency

So, 3780 Hz + (2* 140) = 4060 Hz

And for part (d) we find it's harmonic number by dividing it by the 1st harmonic frequency.

4060 / 140 = 29

It's harmonic number is 29.