A sound source S and a detector D are located at some distance from...

Let the distance between the detector $D$ and the cardboard be $x$.

The sound waves reaching the detector consist of the direct wave from the source $S$ and the reflected wave from the cardboard. The reflected wave travels an additional distance from the detector to the cardboard and back. Thus, the path difference between the direct and reflected waves is:
$\Delta p = 2x$

When the cardboard is displaced by a distance $\Delta x = 20 \text{ cm}$, the new path difference becomes $2(x + \Delta x)$. The change in the path difference is:
$\Delta p' = 2\Delta x = 2 \times 20 \text{ cm} = 40 \text{ cm}$

For the intensity at the detector to vary from a maximum (constructive interference) to an adjacent minimum (destructive interference), the change in path difference must be equal to half of the wavelength $\left(\dfrac{\lambda}{2}\right)$.

Equating the two expressions for the change in path difference:
$\dfrac{\lambda}{2} = 40 \text{ cm}$
$\Rightarrow \lambda = 80 \text{ cm} = 0.8 \text{ m}$

The frequency of the emitted sound $f$ can be calculated using the wave speed equation $v = f\lambda$:
$f = \dfrac{v}{\lambda} = \dfrac{336 \text{ m s}^{-1}}{0.8 \text{ m}} = 420 \text{ Hz}$