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One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. How long is the tube? The speed of sound in air is 343 m/s. One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. How long is the tube? The speed of sound in air is 343 m/s. 1.00 m 2.68 m 1.34 m 0.335 m 0.670 m

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skyluke89

Answer:

1.34 m

Explanation:

For an open-end tube, the frequency difference between two consecutive harmonics is equal to the fundamental frequency of the tube:

[tex]f_1 = f_{n+1}-f_n[/tex]

In this case, we have

[tex]f_{n+1}=576 Hz\\f_n = 448 Hz[/tex]

so, the fundamental frequency is

[tex]f_1=576 Hz-448 Hz= 128 Hz[/tex]

For an open-end tube, the fundamental frequency is also given by:

[tex]f_1 = \frac{v}{2L}[/tex]

where v is the speed of sound and L the length of the tube.

Since we know v = 343 m/s, we can solve the formula for L:

[tex]L=\frac{v}{2f_1}=\frac{343 m/s}{2(128 Hz)}=1.34 m[/tex]

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