A person driving car at a constant speed of 15 m s$$^{-1}$$ is approaching a vertical wall. The person notices a change of 40 Hz in the frequency of his car's horn upon reflection from the wall. The frequency of horn is ______ Hz.
(Given: Speed of sound: 330 m s$$^{-1}$$)

A person driving at $$v_s = 15$$ m/s toward a vertical wall hears a change of $$\Delta f = 40$$ Hz in the frequency of the car horn upon reflection. We need the original frequency, given speed of sound $$v = 330$$ m/s.

This involves two successive Doppler shifts. First, the horn sound travels from the moving car to the stationary wall. The wall receives frequency:

$$f' = f \cdot \frac{v}{v - v_s} = f \cdot \frac{330}{315}$$

(since the source moves toward the stationary observer).

The wall then reflects this sound at frequency $$f'$$, acting as a stationary source, while the driver moves toward it. So the driver hears:

$$f'' = f' \cdot \frac{v + v_o}{v} = f' \cdot \frac{345}{330}$$

Combining both shifts:

$$f'' = f \cdot \frac{330}{315} \cdot \frac{345}{330} = f \cdot \frac{345}{315}$$

Now, the change in frequency is:

$$\Delta f = f'' - f = f\left(\frac{345}{315} - 1\right) = f \cdot \frac{2v_s}{v - v_s} = f \cdot \frac{30}{315}$$

Solving for $$f$$:

$$f = \Delta f \cdot \frac{v - v_s}{2v_s} = 40 \times \frac{315}{30} = 40 \times 10.5 = 420 \text{ Hz}$$

So, the answer is $$420$$ Hz.