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liq [111]
3 years ago
8

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
Physics
1 answer:
mariarad [96]3 years ago
6 0

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:

f_1 = f_{n+1}-f_n

In this case, we have

f_{n+1}=576 Hz\\f_n = 448 Hz

so, the fundamental frequency is

f_1=576 Hz-448 Hz= 128 Hz

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

f_1 = \frac{v}{2L}

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:

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

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