In Quincke's experiment, the sound intensity exhibits a minimum...

In Quincke's tube experiment, when the sliding tube is pulled out by a distance $\Delta x$, the path difference between the two interfering sound waves changes by $\Delta p = 2\Delta x$.

The change in path difference required to transition from a minimum intensity to the adjacent maximum intensity is $\dfrac{\lambda}{2}$.

Equating the two expressions for the change in path difference:

$2\Delta x = \dfrac{\lambda}{2}$

$\Rightarrow \lambda = 4\Delta x$

Given $\Delta x = 16.5 \text{ mm} = 16.5 \times 10^{-3} \text{ m}$, the wavelength is:

$\lambda = 4 \times 16.5 \times 10^{-3} \text{ m} = 66 \times 10^{-3} \text{ m}$

The frequency of the sound source is given by:

$f = \dfrac{v}{\lambda}$

Substituting $v = 330 \text{ m s}^{-1}$:

$f = \dfrac{330}{66 \times 10^{-3}} = 5000 \text{ Hz} = 5.0 \text{ kHz}$