1. Idea of the problem

A stationary source emits sound of frequency $f_0=492\,\text{Hz}$. A car is approaching the source and reflects the sound.

Because the car is moving, there are two Doppler shifts:

• first: the moving car acts as an observer receiving the source sound,
• second: the reflected sound comes back with the car effectively acting as a moving source.

The source then hears:

• the original frequency $f_0=492\,\text{Hz}$,
• the reflected frequency $f_r$.

The beat frequency is $f_b=|f_r-f_0|.$

2. Frequency received by the approaching car

For an observer approaching a stationary source, $f_1=f_0\left(\frac{v+u}{v}\right),$ where

• $v=330\,\text{m/s}$ is speed of sound,
• $u=2\,\text{m/s}$ is speed of the car.

So, $f_1=492\left(\frac{330+2}{330}\right)=492\left(\frac{332}{330}\right).$

3. Reflected frequency heard back at the source

The car reflects the sound at the frequency it receives, so it behaves like a moving source of frequency $f_1$ approaching the stationary source.

For a source approaching a stationary observer, $f_r=f_1\left(\frac{v}{v-u}\right).$

Thus, $f_r=492\left(\frac{332}{330}\right)\left(\frac{330}{328}\right).$

This simplifies to $f_r=492\left(\frac{332}{328}\right)=492\left(\frac{83}{82}\right).$

Now, $492=82\times 6,$ so $f_r=82\times 6\times \frac{83}{82}=6\times 83=498\,\text{Hz}.$

4. Beat frequency

$f_b=|498-492|=6\,\text{Hz}.$

5. Final answer

The beat frequency is $\boxed{6}$ Hz.