91Ó°ÊÓ

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,$n=1,3,5…$

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:

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

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

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$150\text{Hz}$ .