Animated Solution for Physics - Waves: A source of sound S is moving with a velocity of $50\text{ m/s}$ towards a stationary observer. The observer measures the frequency of the source as $1000\text{ Hz}$. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take, velocity of sound in air is $350\text{ m/s}$)

The Doppler effect is one of those beautiful physics phenomena that we experience in everyday life. Think about an ambulance rushing past you on the street. As it approaches, the siren sounds high-pitched and urgent. The moment it passes you, the pitch drops noticeably. This problem is a perfect mathematical translation of that exact experience.

The Approaching Source

Imagine you are standing still on a road, and a car with its siren blaring is rushing towards you at $50\text{ m/s}$. The sound you hear has an apparent frequency of $1000\text{ Hz}$. Because the source is moving towards you, it is essentially "chasing" its own sound waves. This compresses the waves, leading to a higher frequency reaching your ears.

The formula for the apparent frequency when a source approaches a stationary observer is:

Notice the minus sign in the denominator. Subtracting the source velocity $v_s$ from the speed of sound $v$ makes the denominator smaller, which correctly results in a larger apparent frequency $f_{app}$.

Finding the True Voice

Before we can figure out what happens when the car drives away, we need to know the actual frequency $f_{actual}$ emitted by the siren. Let's plug in our known values:

Simplifying the denominator gives us $300$. Rearranging the equation to solve for $f_{actual}$:

Pro Tip: Resist the urge to convert this into a decimal! Keeping it as a fraction will make our final calculation much cleaner and prevent rounding errors.

The Receding Source

Now, the car zooms past you and starts moving away at the same speed of $50\text{ m/s}$. The sound waves are now being stretched out because the source is moving away from the waves it emits backwards. This means you will hear a lower frequency.

For a receding source, the Doppler formula changes slightly:

Here, the plus sign in the denominator increases its value, which correctly models the drop in the apparent frequency.

The Final Calculation

Let's substitute our actual frequency $f_{actual}=\frac{6000}{7}\text{ Hz}$ into this new equation:

The fraction $\frac{350}{400}$ simplifies beautifully to $\frac{7}{8}$. Let's see what happens when we multiply:

The $7$s cancel out perfectly, leaving us with:

And there we have it! The frequency drops from $1000\text{ Hz}$ as it approaches to $750\text{ Hz}$ as it moves away. This elegant cancellation is exactly why we kept the intermediate frequency as a fraction.