One of the harmonic frequencies of tube A with two open ends is 1152 Hz. The nex

Question

One of the harmonic frequencies of tube A with two open ends is 1152 Hz. The next-highest harmonic frequency is 1224 Hz. (a)What harmonic frequency is next highest after the harmonic frequency 288 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 4185 Hz. The next-highest harmonic frequency is 4455 Hz. (c)What harmonic frequency is next highest after the harmonic frequency 3645 Hz? (d) What is the number of this next-highest harmonic?

Explanation / Answer

:  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:

1224Hz - 1152 Hz =72 Hz

Great, we have the first harmonic (72 Hz). So to answer part a)

288 Hz + the 1st harmonic (72 Hz) = the next harmonic (360 Hz)
The number of the next harmonic is pretty simple.

The harmonic frequency / the 1st harmonic frequency = the Harmonic number
So:
360 / 72 =5


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 = (4455 - 4185)/2 = 135 Hz

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

So, 3645 Hz + (2* 135) = 3915 Hz

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

3915 / 135 = 29

It's harmonic number is 29

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