A driver in a car, approaching a vertical wall notices that the frequency of his car horn has changed from $$440$$ Hz to $$480$$ Hz, when it gets reflected from the wall. If the speed of sound in air is $$345\,\text{m s}^{-1}$$, then the speed of the car is:

We begin by noting that the horn of the car emits its natural frequency $$f = 440\ \text{Hz}$$. Because the car is moving towards the wall with an unknown speed $$v$$, the sound waves produced by the horn get compressed. The standard Doppler-effect formula for the frequency $$f_1$$ heard by a stationary observer in front of a source moving with speed $$v$$ is stated as

$$f_1 = \frac{v_s}{\,v_s - v\,}\;f,$$

where $$v_s = 345\ \text{m s}^{-1}$$ is the speed of sound in air.

These waves strike the stationary wall and are reflected. A stationary reflector does not change the frequency of the wave it sends back; it merely acts as a new source emitting the same frequency that it receives. Hence, the wall becomes an effective source of sound of frequency $$f_1$$.

Now the driver (who is moving with the car and therefore with speed $$v$$) hears this reflected sound while approaching the wall. For an observer moving towards a stationary source, the Doppler-effect formula is stated as

$$f_2 = \frac{v_s + v}{v_s}\;f_1,$$

where $$f_2$$ is the frequency finally heard by the driver. Substituting $$f_1$$ from the first result, we get

$$f_2 = \frac{v_s + v}{v_s}\;\left(\frac{v_s}{\,v_s - v\,}\;f\right) = f\,\frac{v_s + v}{v_s - v}.$$

The driver actually hears $$f_2 = 480\ \text{Hz}$$. Therefore,

$$\frac{480}{440} = \frac{v_s + v}{v_s - v}.$$

Simplifying the left-hand side,

$$\frac{480}{440} = \frac{48}{44} = \frac{12}{11}.$$

So we have

$$\frac{12}{11} = \frac{345 + v}{345 - v}.$$

Cross-multiplying gives

$$11\,(345 + v) = 12\,(345 - v).$$

Expanding both sides,

$$3795 + 11v = 4140 - 12v.$$

Bringing the velocity terms to one side and constants to the other,

$$11v + 12v = 4140 - 3795,$$

$$23v = 345.$$

Hence,

$$v = \frac{345}{23} = 15\ \text{m s}^{-1}.$$

To convert this speed into kilometres per hour, we use the relation $$1\ \text{m s}^{-1} = \frac{18}{5}\ \text{km h}^{-1}$$, so

$$v = 15 \times \frac{18}{5} = 54\ \text{km h}^{-1}.$$

Hence, the correct answer is Option A.