(a) 288 Hz
The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, [tex]f_1[/tex] (first harmonic):
[tex]f_{n+1}-f_n = f_1[/tex] (1)
In this problem, we are told the frequencies of two successive harmonics:
[tex]f_n = 576 Hz\\f_{n+1}=648 Hz[/tex]
So the fundamental frequency is:
[tex]f_1 = 648 Hz-576 Hz=72 Hz[/tex]
Now we know that one of the the harmonics is [tex]f_n=216 Hz[/tex], so its next highest harmonic will have a frequency of
[tex]f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz[/tex]
(b) n=4
The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:
[tex]f_n=n f_1[/tex] (2)
Since we know [tex]f_n = 288 Hz[/tex], we can solve (2) to find the number n of this harmonic:
[tex]n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4[/tex]
(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:
[tex]f_n=(2n+1) f_1[/tex] (2)
so, the difference between any two harmonics tube is equal to:
[tex]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[/tex] (1)
In this problem, we are told the frequencies of two successive harmonics:
[tex]f_n = 4699 Hz\\f_{n+1}=4953 Hz[/tex]
So, according to (1), the fundamental frequency is equal to half of this difference:
[tex]f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz[/tex]
Now we know that one of the harmonics is [tex]f_n=4191 Hz[/tex], so its next highest harmonic will have a frequency of
[tex]f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz[/tex]
(d) n=17
We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:
[tex]f_n=(2n+1) f_1[/tex] (2)
Since we know [tex]f_n = 4445 Hz[/tex], we can solve (2) to find the number n of this harmonic:
[tex]n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17[/tex]
