For a particular tube, here are four of the six harmonic frequencies below 1000Hz : 300, 600 , 750 , and 900Hz. What two frequencies are missing from the list?

The four of the six harmonic frequencies below 1000 Hz are 300Hz, 600Hz, 750Hz and 900Hz.

By using the equation for the harmonic frequencies f, find the ratio v/L. By using this value for v/L , find the two missingharmonic frequencies below 1000Hz.

Formula is as follow:

Ifthetube is open at both ends, the harmonic frequencies are,

$f=\frac{v}{\lambda }=\frac{nv}{2L}$

Where,

f is frequency, L is length and v is velocity

If the tube is open at both ends, the resonant frequencies are,

$f=\frac{v}{\lambda }=\frac{nv}{2L}$

Where,

v is the speed of sound, $\lambda$is wavelength, L is length of tube and $n=1,2,3,\dots$

Let, $f_6=900Hz$ is the $6^{th}$harmonic frequency.

$f_6=\frac{6v}{2L}$

$900=\frac{6v}{2L}$

$\left(\frac{v}{L}\right)=\frac{900}{3}$

$\left(\frac{v}{L}\right)=300$

Now, find the two missing resonant frequencies.

The first harmonic frequency $f_1$,

$f_1=\frac{1}{2}\left(\frac{v}{L}\right)$

$f_1=\frac{1}{2}\times 300$

$f_1=150Hz$

Similarly, the third harmonic frequency $f_3$,

$f_3=\frac{3}{2}\left(\frac{v}{L}\right)$

$f_3=\frac{3}{2}\times 300$

$f_3=450Hz$

Thus, by using the equation for the harmonic frequencies f , this question can be solved.