Animated Solution for Physics - Waves: The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes from $420\text{ Hz}$ to $490\text{ Hz}$, when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is $330{\text{ ms}}^{-1}$.

Mastering the Doppler Effect

Imagine you are driving a bus towards a massive, unyielding wall. You press the horn. The sound travels forward, hits the wall, and bounces back to you. But something fascinating happens—the pitch of the echo you hear is noticeably higher than the original honk! This isn't magic; it's the Doppler Effect in action.

In this problem, we are going to decode this exact scenario. We will break down the journey of the sound wave into two distinct phases. By the end of this, you won't just know the formula; you'll feel the physics behind it.

The Journey to the Wall

Let's freeze time right after the horn is pressed. The bus is moving towards the wall with an unknown speed, let's call it $v_B$. The horn emits a sound wave with a frequency of $f_0=420\text{ Hz}$.

Because the bus is chasing its own sound waves, the wavefronts get squished together in front of the bus. To the stationary wall, these squished waves arrive more frequently. The wall "hears" a higher frequency, $f_1$.

Using the Doppler effect formula for a moving source and a stationary observer, we write:

Substituting our known values (where the speed of sound $v=330\text{ m/s}$):

We don't know $v_B$ yet, so we'll keep this equation safe in our back pocket.

The Echo Returns

Now, the sound hits the wall. The wall acts like a perfect mirror for the sound, becoming a new, stationary source that emits the exact frequency it just received: $f_1$.

But the story isn't over. The bus is still moving forward! Now, the bus is a moving observer rushing towards this stationary source. Because the driver is driving into the oncoming sound waves, they hit his ears even faster. The final frequency he hears, $f_2$, is given by:

Let's bring our two phases together. We substitute our expression for $f_1$ into the $f_2$ equation:

Notice how beautifully the $330$ in the numerator and denominator cancel out! We are given that the final echo heard by the driver is $f_2=490\text{ Hz}$. Plugging this in, we get:

Dividing both sides by $420$, we simplify the ratio:

Componendo and Dividendo

We could cross-multiply here, but let's use a powerful algebraic weapon: the Componendo and Dividendo rule. If $\frac{a}{b}=\frac{c}{d}$, then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.

Applying this to our equation:

This simplifies incredibly fast:

Units!

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

Rounding to the nearest integer, we get $91\text{ km/h}$. The driver is certainly speeding, but more importantly, you just flawlessly navigated a double-Doppler effect problem!