Question

In Kundt's tube experiment, a brass rod of length $1\text{m}$ is used and the average length of loop obtained in air is $10.3\text{cm}$ and in $C{O}_2$ is $8\text{cm}$. Calculate the speed of sound in brass and in $C{O}_2$.
Given, speed of sound in air is $350\text{m/s}$.

• A. $5000\text{m/s},425.8\text{m/s}$
• B. $3400\text{m/s},271.8\text{m/s}$
• C. $2250\text{m/s},133.8\text{m/s}$
• D. $4800\text{m/s},350.8\text{m/s}$

Answer 1

The correct option is B $3400\text{m/s},271.8\text{m/s}$

Given that
Length of brass rod $=1\text{m}$
Since rod is clamped at the middle, wavelength of sound wave in rod is given by
$\frac{{\lambda }_r}{2}=1\Rightarrow {\lambda }_r=2\text{m}$

In air,
$\frac{{\lambda }_a}{2}=10.3\text{cm}=0.103\text{m}$
$\Rightarrow {\lambda }_a=0.206\text{m}$
In $C{O}_2$,
$\frac{{\lambda }_{C{O}_2}}{2}=8\text{cm}=8\times {10}^{-2}\text{m}$
$\Rightarrow {\lambda }_{C{O}_2}=16\times {10}^{-2}\text{m}$

Since sound wave is produced at brass rod and it propagates through air and $C{O}_2$ medium,
$\therefore$ Frequency of wave in each medium will be same.
$\Rightarrow f_{\text{rod}}=f_{\text{air}}=f_{C{O}_2}$

$\Rightarrow \frac{v_{\text{rod}}}{{\lambda }_r}=\frac{v_{\text{air}}}{{\lambda }_a}=\frac{v_{C{O}_2}}{{\lambda }_{C{O}_2}}$

$\therefore \frac{v_{\text{rod}}}{{\lambda }_r}=\frac{v_{\text{air}}}{{\lambda }_a}$

$\Rightarrow v_{\text{rod}}=v_{\text{air}}\times \frac{{\lambda }_r}{{\lambda }_a}=\frac{350\times 2}{0.206}\approx 3400\text{m/s}$

And,
$\frac{v_{\text{air}}}{{\lambda }_a}=\frac{v_{C{O}_2}}{{\lambda }_{C{O}_2}}$
$\Rightarrow v_{C{O}_2}=v_{\text{air}}\times \frac{{\lambda }_{C{O}_2}}{{\lambda }_a}=350\times \frac{0.16}{0.206}=271.8\text{m/s}$