Question

In a Quincke's tube experiment, the sound intensity being detected at an appropriate point, changes from minimum to maximum for the second time, when the sliding tube is drawn out by $18.0\text{cm}.$ If the speed of sound in air is $340\text{m/s}$, then find the frequency of this sounding source.

• A. $1417\text{Hz}$
• B. $2833\text{Hz}$
• C. $1889\text{Hz}$
• D. $944\text{Hz}$

Answer 1

The correct option is A $1417\text{Hz}$

Since sliding tube is pulled out by $18\text{cm}$,
Path difference, $\Delta x=2\times 18=36\text{cm}.........\left(1\right)$

Initially, the position of the tube corresponds to the minimum intensity. When maximum intensity is heard for the first time, the path difference increases by $\frac{\lambda }{2}$ and when heard for the second time, there is an additional path difference of $\lambda$.
So, Total path difference,$\Delta x=\frac{\lambda }{2}+\lambda =\frac{3\lambda }{2}........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{3\lambda }{2}=36\Rightarrow \lambda =24\text{cm}$
So, frequency of sound emitted by source will be
$f=\frac{v}{\lambda }=\frac{340}{24\times {10}^{-2}}=1416.66\text{Hz}\approx 1417\text{Hz}$