Animated Solution for Physics - Waves: A stationary source emits sound waves of frequency $500\text{ Hz}$. Two observers moving along a line passing through the source detect sound to be of frequencies $480\text{ Hz}$ and $530\text{ Hz}$. Their respective speeds are in ${\text{ms}}^{-1}$, (Take, speed of sound $=300\text{ m/s}$)

Have you ever noticed how the pitch of an ambulance siren drops the moment it passes you? That is the Doppler Effect in action! In this problem, we are going to use this exact principle to play detective and figure out how fast two observers are moving just by listening to the frequencies they hear.

Imagine a stationary sound source blasting a continuous tone at $500\text{ Hz}$. This is our true frequency, $f_0$.

Now, we have two observers moving along a straight line that passes through this source.

Observer 1 hears a frequency of $480\text{ Hz}$. Since $480\text{ Hz}$ is less than the true $500\text{ Hz}$, the sound waves are being "stretched out" relative to them. This immediately tells us that Observer 1 is moving away from the source.

Observer 2, on the other hand, hears a frequency of $530\text{ Hz}$. Because $530\text{ Hz}$ is greater than $500\text{ Hz}$, the sound waves are being "compressed" relative to them. This means Observer 2 is moving towards the source.

The Doppler effect formula for a stationary source and a moving observer is given by:

Here, $v$ is the speed of sound ($300\text{ m/s}$), and $v_0$ is the speed of the observer.

The sign in the numerator depends on the direction of motion: - We use a minus sign ($-$) when the observer moves away from the source. - We use a plus sign ($+$) when the observer moves towards the source.

Let's set up the equation for Observer 1, who is moving away. We will call their speed $v_1$.

To solve this, let's isolate the term with $v_1$:

Simplifying the left side, we get $48\times 6$, which is $288$.

Solving for $v_1$, we find:

Now, let's do the same for Observer 2, who is moving towards the source. We will call their speed $v_2$ and use the plus sign.

Isolating the term with $v_2$:

Simplifying the left side, we get $53\times 6$, which is $318$.

Solving for $v_2$, we find:

We have successfully calculated the speeds of both observers! Observer 1 is moving at $12\text{ m/s}$, and Observer 2 is moving at $18\text{ m/s}$.

Looking at our options, this perfectly matches option (b). The math confirms our physical intuition flawlessly!