Animated Solution for Physics - Waves: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency $10\text{ GHz}$. What is the frequency of the microwave measured by the observer? (speed of light $=3\times 10^8{\text{ ms}}^{-1}$)

Imagine you are an observer in a spaceship, zooming towards a stationary microwave source at a mind-bending speed—exactly half the speed of light! The source is emitting microwaves at a frequency of $10\text{ GHz}$. Our mission is to find out what frequency you will measure from your incredibly fast spaceship.

When we see a problem involving a moving observer and a stationary source, our first instinct is to grab the classical Doppler effect formula. For an observer moving towards a stationary source, the classical formula is $f_{\text{observed}}=f_{\text{actual}}\left(1+\frac{v}{c}\right)$.

If we blindly plug our values into this classical equation, we get $f_{\text{observed}}=10\left(1+0.5\right)=15\text{ GHz}$. Look at the options! Option (c) is exactly $15.3\text{ GHz}$ (close enough to trick you). This is a classic trap set by the examiners.

Why does the classical formula fail here? Because your speed is $v=\frac{c}{2}$, which is highly comparable to the speed of light. At such extreme velocities, the universe behaves differently. Time dilation comes into play, and we must use the principles of Special Relativity.

To solve this correctly, we must use the Relativistic Doppler Effect formula. Unlike the classical Doppler effect, which depends on whether the source or the observer is moving relative to the medium, the relativistic version only depends on the relative velocity between the source and the observer.

For an observer and source approaching each other, the relativistic Doppler formula is:

This beautiful equation accounts for both the classical bunching of waves and the relativistic time dilation experienced by the moving observer.

Now, let's carefully substitute our given values into the master equation. We know the actual frequency $f_{\text{actual}}=10\text{ GHz}$ and the velocity ratio $\frac{v}{c}=\frac{1}{2}$.

Let's simplify the fraction inside the square root. The numerator becomes $1+\frac{1}{2}=\frac{3}{2}$, and the denominator becomes $1-\frac{1}{2}=\frac{1}{2}$.

The halves in the numerator and denominator cancel out perfectly, leaving us with a very clean expression:

We know that the value of $\sqrt{3}$ is approximately $1.732$. Multiplying this by $10$, we get:

Rounding to one decimal place, we get $17.3\text{ GHz}$. This matches perfectly with option (b). By recognizing the need for relativity, we successfully avoided the trap and arrived at the exact correct answer!