91Ó°ÊÓ

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,

$f=\frac{v}{\mathrm{λ}}=\frac{nv}{2L}$

Where,

f is frequency, L is length and v is velocity

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

$f=\frac{v}{\mathrm{λ}}=\frac{nv}{2L}$

Where,

v is the speed of sound, $\mathrm{λ}$is wavelength, L is length of tube and $n=1,2,3,…$

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 $f_1$,

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

$f_1=\frac{1}{2}×300$

$f_1=150Hz$

Similarly, the third harmonic frequency $f_3$,

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

$f_3=\frac{3}{2}×300$

$f_3=450Hz$

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