During Quincke's experiment, the detected sound transitions from a...

In Quincke's tube experiment, when the sliding tube is moved by a distance $x$, the path difference between the two interfering sound waves changes by $2x$.

For the sound intensity to transition from a maximum to a minimum, the change in path difference must be $\dfrac{\lambda}{2}$.

Equating the two expressions for the change in path difference:

$2x = \dfrac{\lambda}{2} \Rightarrow \lambda = 4x$

Given $x = 2.50 \text{ cm} = 0.025 \text{ m}$, the wavelength is:

$\lambda = 4 \times 0.025 = 0.1 \text{ m}$

The frequency of the sound wave is given by $f = \dfrac{v}{\lambda}$.

Substituting $v = 340 \text{ m s}^{-1}$ and $\lambda = 0.1 \text{ m}$:

$f = \dfrac{340}{0.1} = 3400 \text{ Hz} = 3.4 \text{ kHz}$