Question

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

Answer 1

1.34 m

Step-by-step 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 = (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=(v)/(2f_1)=(343 m/s)/(2(128 Hz))=1.34 m`