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

Imagine you are driving a car towards a large vertical wall. You press the horn, which emits a sound of $440\text{ Hz}$. The sound waves travel to the wall, bounce back, and when you hear them, the pitch has shifted up to $480\text{ Hz}$. This is a classic example of the Doppler effect involving an echo. Let's break down the journey of these sound waves.

To solve this, we must realize that the Doppler shift happens twice. First, the car acts as a moving source sending sound to the stationary wall. Second, the wall acts as a stationary source reflecting that exact sound back to the moving car. We will apply the Doppler formula step by step.

Let's look at the first stage. The car is moving towards the wall with speed $v_c$. The wall is stationary. According to the Doppler effect, the frequency $f_1$ received by the wall will be higher than $f_0$. The formula is:

Now for the second stage. The wall reflects the sound, so it becomes a stationary source emitting frequency $f_1$. The car is now the observer, moving towards this source. The frequency $f_2$ heard by the driver will be even higher. The formula becomes:

Let's combine these two stages into a single master equation. By substituting the expression for $f_1$ into our second equation, the speed of sound $v$ in the numerator and denominator beautifully cancels out. We are left with a direct relationship between the final heard frequency $f_2$ and the original frequency $f_0$:

Now, let's plug in the numbers given in the problem. The original frequency is $440\text{ Hz}$, the final is $480\text{ Hz}$, and the speed of sound is $345\text{ m/s}$. We get:

Simplifying the ratio $480$ over $440$ gives us $12$ over $11$. Notice how clean the numbers are becoming. Don't rush the calculation here.

Let's solve for $v_c$. Cross-multiplying gives us:

Instead of multiplying $12$ and $345$, let's be smart. Bring the $345$ terms to one side and the $v_c$ terms to the other. We get:

Dividing $345$ by $23$ gives us exactly $15\text{ m/s}$.

We have the speed as $15\text{ m/s}$, but look at the options! They are all in $\text{km/h}$. This is a classic trap. To convert $\text{m/s}$ to $\text{km/h}$, we multiply by $\frac{18}{5}$.

So, the car is moving at $54\text{ km/h}$. Option (a) is the correct answer.