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?

(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:

inverse v=i*r

nice nice

the answer you are looking for is d. there are no holes for blood to enter or leave the system. your welcome