A pipe$\mathbf{0}\mathbf{.60}\mathbf{}\mathbf{\text{m}}$ long and closed at one end is filled with an unknown gas .The third lowest harmonic frequency for the pipe is $\mathbf{\text{750Hz}}$. (a) What is the speed of sound in the unknown gas? (b) What is the fundamental frequency for this pipe when it is filled with the unknown gas?

1. The length of pipe is $L=0.60\text{m}$
2. The third lowest harmonic frequency for the pipe is$f_5=750\text{Hz}$

Use the concept of one open-ended pipe. Only odd harmonic can be set up when the pipe is open at one end. The third lowest harmonic for the pipe is n=5. We can use the relation between the fundamental and third-lowest harmonic frequency to find the fundamental frequency of the pipe closed at one end.

Formulae:

Resonant frequencies for the pipe open at one end, $f=\frac{nv}{4L}$ (i)

For,$n=1,3,5\dots$

The third lowest frequency of this harmonic motion, $f_5=5f$ (ii)

The third lowest harmonic is$n=5$with their resonant frequency using equations (i) and (ii) is given as:

$f_5=\frac{5v}{4L}$

So, the speed of the sound in the unknown gas is using above equation is given by:

$\begin{array}{rcl}v& =& \frac{f_5\times 4L}{5}\\ & =& \frac{750\text{Hz}\times 4\times 0.60\text{m}}{5}\\ & =& 3.6\times {10}^2\frac{\text{m}}{\text{s}}\\ & & \end{array}$

Hence, the speed of the sound in the gas is$3.6\times {10}^2\frac{\text{m}}{\text{s}}$

The relation between the fundamental frequency and the third lowest harmonic frequency for the pipe when it is filled with the unknown gas is given by equation (ii), which is:

$\begin{array}{c}f=\frac{f_5}{5}\\ =\frac{750\text{Hz}}{5}\\ =150\text{Hz}\end{array}$

Hence, the value of fundamental frequency is$150\text{Hz}$ .