Question

In Quincke’s experiment, the sound intensity has a minimum value $I$at a particular position. As the sliding tube is pulled out by a distance of $16.5mm$, the intensity increases to a maximum of $9I$. Take the speed of sound in air to be $330\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

(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

Step 1: Give Information

Distance$=16.5mm$

Max intensity$=9I$

Speed of sound in air$=330\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2: Formula

$\mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }}\mathrm{where}\mathrm{f}\mathrm{is}\mathrm{frequency},\mathrm{v}\mathrm{is}\mathrm{velocity}\mathrm{of}\mathrm{wave},\mathrm{\lambda }\mathrm{is}\mathrm{wavelength}$

Step 3: (a) Find the frequency of the sound source

The distance between the maximum to the minimum intensity in terms of the wavelength and the distance is,

$\Rightarrow \frac{\lambda }{4}=16.5mm$

$\Rightarrow \lambda =16.5x4$

$\Rightarrow \lambda =66mm$

$\Rightarrow \lambda =66x{10}^{-3}m$

Know,$v=f\lambda$, Where, $v$= Velocity, $f$= Frequency and $\lambda$= Wavelength

$orf=\frac{v}{\lambda }=\frac{340}{[66\times {10}^{-3}]}=5kHz$

Step 3: (b) Ratio of maximum intensity to minimum intensity

Find the ratio of maximum intensity to minimum intensity.

$\Rightarrow \frac{(A_1+A_2)}{(A_1–A_2)}=\frac{1}{9}$

$\Rightarrow \frac{A_1}{A_2}=\frac{2}{1}$,Where, $A_{1}and{A}_2$ is amplitude.

The ratio of the amplitudes is 2:1.

Hence, the answer for (a) and (b) is $5kHz$ and $2:1$ respectively.