Question

In Quincke's experiment, the sound intensity has a minimum value l at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm, the intensity increases to a maximum of 9 l. Take the speed of sound in air to be 330 m s⁻¹. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.

Answer 1

The sliding tube is pulled out by a distance of 16.5 mm.
Speed of sound in air, v = 330 ms⁻¹

(a) As per the question, we have:
$$\begin{gathered} \frac{\mathrm{\lambda }}{4}=16.5\mathrm{mm} \\ \Rightarrow \mathrm{\lambda }=16.5\times 4=66\mathrm{mm}=66\times {10}^{-3}\mathrm{m} \end{gathered}$$

We know:
v = f $\lambda$
$\therefore f=\frac{v}{\lambda }$
$\Rightarrow f=\frac{v}{\mathrm{\lambda }}=\frac{340}{66\times {10}^{-3}}=5\mathrm{kHz}$

(b)​
Ratio of maximum intensity to minimum intensity:
$$\begin{gathered} \frac{I_{Max}}{I_{Min}}=\frac{K{\left(A_1-A_2\right)}^2}{K{\left(A_1+A_2\right)}^2}=\frac{I}{9I} \\ \Rightarrow \frac{{\left(A_1-A_2\right)}^2}{{\left(A_1+A_2\right)}^2}=\frac{1}{9} \\ \mathrm{Taking}\mathrm{square}\mathrm{roots}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \end{gathered}$$
$$\begin{gathered} \frac{A_1+A_2}{A_1-A_2}=\frac{3}{1} \\ \Rightarrow \frac{A_1}{A_2}=\frac{3+1}{3-1}=\frac{2}{1} \end{gathered}$$
So, the ratio of the amplitudes is 2.