For the stationary observer:

Frequency of the sound produced by the whistle, ν = 400 Hz

Speed of sound = 340 m/s

Velocity of the wind, v = 10 m/s

As there is no relative motion between the source and the observer, the observer will hear the same frequency of the sound as produced by the source, i.e., 400 Hz.

The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units.

$\mathrm{Effective} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{sound}, {\mathrm{v}}_{\mathrm{e}} = 340 + 10 = 350 \mathrm{m}/\mathrm{s}$
$\mathrm{The} \mathrm{wavelength} \left(\mathrm{\lambda }\right) \mathrm{of} \mathrm{the} \mathrm{sound} \mathrm{heard} \mathrm{by} \mathrm{the} \mathrm{observer}:$
$\mathrm{\lambda }=\frac{{\mathrm{v}}_{\mathrm{e}}}{\mathrm{\nu }}=\frac{350}{400}=0.875 \mathrm{m}$
$$
$\mathrm{For} \mathrm{the} \mathrm{running} \mathrm{observer}:$
$\mathrm{Velocity} \mathrm{of} \mathrm{the} \mathrm{observer}, {\mathrm{v}}_{\mathrm{o}} = 10 \mathrm{m}/\mathrm{s}$
$\mathrm{The} \mathrm{apparent} \mathrm{frequency} \mathrm{heard} \mathrm{by} \mathrm{the} \mathrm{observer}:$
$\begin{array}{rl}\mathrm{v}\text{'}& =\left(\frac{\mathrm{v}+{\mathrm{v}}_0}{\mathrm{v}}\right)\mathrm{v}\end{array}=\left(\frac{340+10}{340}\right)\times 400=411.76\mathrm{Hz}$

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s
The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m. Hence, the given two situations are not exactly identical.