Question

A uniform narrow tube 1.70 m long is open at both ends. it resonates at two successive harmonics of frequency 280 hz and 350 hz. what is the speed of sound in the gas in the tube?

Answer 1

Let's call
`f_n` the frequency of the nth-harmonic and
`f_(n+1)` the frequency of the (n+1)th harmonic, wish is the next harmonic.
Since the frequency of the nth-harmonic is n times the fundamental frequency f1:

`f_n = n f_1`
then the difference between two successive harmonics is equal to the fundamental frequency of the tube:

`f_(n+1)-f_n = (n+1)f_1 - nf_1 = nf_1 + f_1 - nf _1 = f_1`

so, by using 350 Hz and 280 Hz as successive harmonics, we find the fundamental frequency of the tube:

`f_1 = 350 Hz - 280 Hz = 70 Hz`

The wavelength of the first harmonic is twice the length of the tube:

`\lambda = 2 L=2 \cdot 1.70 m=3.40 m`

And since we know both frequency and wavelength, we can find the speed of the wave in the tube, which is the speed of sound in the gas in the tube:

`v= \lambda f_1 = (3.40 m)(70 Hz)=238 m/s`