Question

A Kundt's tube apparatus has a steel rod of length $1\text{m}$ clamped at the centre. It is vibrated in its fundamental mode at a frequency of $2600\text{Hz}$. The lycopodium powder dispersed in the tube collects into heaps seperates by $6.5\text{cm}$. Calculate the speed of sound in air

• A. $356\text{m/s}$
• B. $338\text{m/s}$
• C. $328\text{m/s}$
• D. $380\text{m/s}$

Answer 1

The correct option is B $338\text{m/s}$

In Kundt's tube experiment, as the rod behaves like an open organ pipe,
Fundamental mode of vibration $f=\frac{v_r}{2{\ell }_r}=\frac{v_r}{{\lambda }_r}$
where $v_r=$ velocity of sound in rod
${\ell }_r=$ length of rod clamped at middle
${\lambda }_r=$ wavelength of sound in rod

From the data given in the question,

$v_r=2600\times 2\times 1=5200\text{m/s}$

Since the frequency of vibration of the rod and air are same, we can say that,

$\frac{v_r}{v_a}=\frac{{\lambda }_r}{{\lambda }_a}=\frac{2{\ell }_r}{2\Delta \ell }$
where $\Delta \ell =$ distance between heaps

$\Rightarrow v_a=v_r\frac{\Delta \ell }{{\ell }_r}$

Substituting the given data,

$v_a=\frac{5200\times 6.5\times {10}^{-2}}{1}=338\text{m/s}$

Thus, option (b) is the correct answer.