Question

In a Quincke's tube experiment, the sound detected is changed from a maximum to minimum when sliding tube is moved through a distance $2\text{cm}$. Find the frequency of sound emitted by the source, if the speed of sound in air is $340\text{m/s}$.

• A. $2428\text{Hz}$
• B. $2125\text{Hz}$
• C. $4250\text{Hz}$
• D. $8500\text{Hz}$

Answer 1

The correct option is C $4250\text{Hz}$

Since sliding tube is moved by $2\text{cm}$,
Path difference, $\Delta x=2\times 2=4\text{cm}.........\left(1\right)$

From the data given in the question, the sound intensity is changed from maximum to minimum, so the path difference will be $\frac{\lambda }{2}$.

So, path difference,$\Delta x=\frac{\lambda }{2}........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{\lambda }{2}=4\Rightarrow \lambda =8\text{cm}$
So, frequency of sound emitted by source will be
$f=\frac{v}{\lambda }=\frac{340}{8\times {10}^{-2}}=4250\text{Hz}$