Animated Solution for Physics - Waves: Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of $20{\text{ ms}}^{-1}$ with respect to the ground. If an observer in car A detects a frequency $2000\text{ Hz}$ of the sound coming from car B, what is the natural frequency of the sound source in car B? (speed of sound in air = $340{\text{ ms}}^{-1}$)

The Doppler Effect is one of those beautiful phenomena in physics that you experience all the time in real life. Think about an ambulance speeding past you—the pitch of the siren drops noticeably as it goes from approaching you to moving away. In this problem, we are going to mathematically decode exactly how that frequency shift happens when both the source and the observer are in motion!

Imagine you are standing on a long, straight highway. Two cars, A and B, are speeding away from each other in opposite directions. Both cars are moving at a brisk $20\text{ m/s}$. Car B is honking its horn, making it the source of our sound. The driver in Car A is listening to this horn, making them the observer.

The sound waves emitted by Car B have to travel through the air to reach Car A. The speed of these sound waves is given as $340\text{ m/s}$. The driver in Car A hears a frequency of $2000\text{ Hz}$. This is the apparent frequency ($f$). Our mission is to find the true frequency ($f_0$) of the horn in Car B.

To solve this, we need our trusty Doppler Effect formula:

Here is the golden rule for getting the signs right: Always think about how the motion affects the frequency.

1. Observer's Motion: Car A is moving away from Car B. Moving away means the observer encounters fewer wave crests per second, which decreases the apparent frequency. To make the overall fraction smaller, we must subtract the observer's velocity in the numerator. So, we use $v-v_o$.

2. Source's Motion: Car B is moving away from Car A. When a source moves away, it stretches the sound waves out (longer wavelength), which also decreases the apparent frequency. To make the overall fraction smaller, we must add the source's velocity in the denominator. So, we use $v+v_s$.

Putting it all together, our specific formula for this scenario is:

Now, let's plug in the numbers. We know $v=340\text{ m/s}$, $v_o=20\text{ m/s}$, $v_s=20\text{ m/s}$, and $f=2000\text{ Hz}$.

Let's simplify the fraction:

We can easily reduce $\frac{320}{360}$ by dividing the numerator and denominator by $40$, which gives us $\frac{8}{9}$.

Finally, we isolate $f_0$ to find the true frequency:

And there we have it! The natural frequency of the sound source in Car B is exactly $2250\text{ Hz}$. Notice how the true frequency is higher than the apparent frequency ($2000\text{ Hz}$). This makes perfect sense because both cars moving away from each other causes a significant "redshift" (drop in pitch) in the sound!