A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i
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(a) 288 Hz
The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, (first harmonic):
(1)
In this problem, we are told the frequencies of two successive harmonics:
So the fundamental frequency is:
Now we know that one of the the harmonics is , so its next highest harmonic will have a frequency of
(b) n=4
The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:
(2)
Since we know , we can solve (2) to find the number n of this harmonic:
(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:
(2)
so, the difference between any two harmonics tube is equal to:
(1)
In this problem, we are told the frequencies of two successive harmonics:
So, according to (1), the fundamental frequency is equal to half of this difference:
Now we know that one of the harmonics is , so its next highest harmonic will have a frequency of
(d) n=17
We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:
(2)
Since we know , we can solve (2) to find the number n of this harmonic: