Question

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What harmonic frequency is next highest after the harmonic frequency 216 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 4699 Hz. The next-highest harmonic frequency is 4953 Hz. (c) What harmonic frequency is next highest after the harmonic frequency 4191 Hz? (d) What is the number of this next-highest harmonic?

Answer 1

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency,
`f_1` (first harmonic):

`f_(n+1)-f_n = f_1` (1)

In this problem, we are told the frequencies of two successive harmonics:

`f_n = 576 Hz\\f_(n+1)=648 Hz`

So the fundamental frequency is:

`f_1 = 648 Hz-576 Hz=72 Hz`

Now we know that one of the the harmonics is
`f_n=216 Hz`, so its next highest harmonic will have a frequency of

`f_(n+1)=f_n+f_1 = 216 Hz+72 Hz=288 Hz`

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

`f_n=n f_1` (2)

Since we know
`f_n = 288 Hz`, we can solve (2) to find the number n of this harmonic:

`n=(f_n)/(f_1)=(288 Hz)/(72 Hz)=4`

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

`f_n=(2n+1) f_1` (2)

so, the difference between any two harmonics tube is equal to:

`f_(n+1)-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1` (1)

In this problem, we are told the frequencies of two successive harmonics:

`f_n = 4699 Hz\\f_(n+1)=4953 Hz`

So, according to (1), the fundamental frequency is equal to half of this difference:

`f_1 = (4953 Hz-4699 Hz)/(2)=127 Hz`

Now we know that one of the harmonics is
`f_n=4191 Hz`, so its next highest harmonic will have a frequency of

`f_(n+1)=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz`

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

`f_n=(2n+1) f_1` (2)

Since we know
`f_n = 4445 Hz`, we can solve (2) to find the number n of this harmonic:

`n=(1)/(2)((f_n)/(f_1)-1)=(1)/(2)((4445 Hz)/(127 Hz)-1)=17`