In a Quincke's experiment, the sound intensity has a minimum valu… - Vidyadip

Question

In a 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 330m/s.

Find the frequency of the sound source.
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

According to the data

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

$\Rightarrow\lambda=66\text{mm}=66\times10^{-3}\text{m}$

$\Rightarrow\text{n}=\frac{\text{V}}{\lambda}=\frac{330}{66\times10^{-3}}=5\text{KHz}$

$\text{I}_\text{minimum}=\text{K}(\text{A}_1-\text{A}_2)^2=\text{I}\Rightarrow\text{A}_1-\text{A}_2=11$

$\text{I}_\text{minimum}=\text{K}(\text{A}_1+\text{A}_2)^2=9\Rightarrow\text{A}_1+\text{A}_2=31$

So, $\frac{\text{A}_1+\text{A}_2}{\text{A}_1+\text{A}_2}=\frac{\text{3}}{4}$

$\Rightarrow\frac{\text{A}_1}{\text{A}_2}=\frac{2}{1}$

So, the ratio amplitudes is 2.

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