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.
A)1.34 m
B)0.670 m
C)0.335 m
D)1.00 m
E)2.68 m

Question

Quizplus · Accepted Answer

The harmonics of a tube open at one end and closed at the other are odd multiples of the fundamental frequency. If one harmonic is 448 Hz and the next higher harmonic is 576 Hz, the difference between them represents two harmonic steps (since even harmonics are absent in such a tube). This means the fundamental frequency (first harmonic) can be found by dividing the difference by 2. The difference is 576 Hz - 448 Hz = 128 Hz. Dividing by 2 gives us 64 Hz for the fundamental frequency. The wavelength of the fundamental frequency ($\lambda_1$) is four times the length of the tube ($L$), because a quarter wavelength fits in the tube. So, $\lambda_1 = 4L$. The speed of sound ($v$) is the product of the frequency ($f$) and the wavelength ($\lambda$), so $v = f\lambda$. Substituting the values for the fundamental frequency and rearranging to solve for $L$, we get $L = \frac{v}{4f} = \frac{343 \, \text{m/s}}{4 \times 64 \, \text{Hz}} = 1.34 \, \text{m}$.

One of the Harmonics of a Column of Air in a Tube