Two sources of sound S$$_1$$ and S$$_2$$ produce sound waves of same frequency 660 Hz. A listener is moving from source S$$_1$$ towards S$$_2$$ with a constant speed u$$_0$$ m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u$$_0$$ equals:

The two sources S$$_1$$ and S$$_2$$ are at rest and emit sound of the same true frequency $$f = 660\ \text{Hz}$$. The listener, however, is moving from S$$_1$$ towards S$$_2$$ with a constant speed $$u_0\ \text{m s}^{-1}$$. Because of this motion, the listener approaches S$$_2$$ and recedes from S$$_1$$ simultaneously, so the apparent (heard) frequencies of the two sources become different. Beats are produced due to the superposition of these two slightly different apparent frequencies.

First, we recall the formula for the apparent frequency when a listener moves and the source remains stationary:

$$f' = f\left(\dfrac{v \pm u}{v}\right)$$

Here $$v = 330\ \text{m s}^{-1}$$ is the speed of sound in air, and the plus sign is used when the listener moves towards the source while the minus sign is used when the listener moves away from it.

For source S$$_2$$, the listener is moving towards it, so the apparent frequency heard is

$$f_2' = f\left(\dfrac{v + u_0}{v}\right).$$

For source S$$_1$$, the listener is moving away from it, so the apparent frequency heard is

$$f_1' = f\left(\dfrac{v - u_0}{v}\right).$$

The beat frequency is the magnitude of the difference of these two apparent frequencies, that is

$$f_{\text{beat}} = \left|\,f_2' - f_1'\,\right|.$$

Substituting the expressions of $$f_2'$$ and $$f_1'$$ we get

$$\begin{aligned} f_{\text{beat}} &= \left|\,f\left(\dfrac{v + u_0}{v}\right) - f\left(\dfrac{v - u_0}{v}\right)\right| \\ &= f \left|\dfrac{(v + u_0) - (v - u_0)}{v}\right| \\ &= f \left|\dfrac{2u_0}{v}\right| \\ &= f\,\dfrac{2u_0}{v}. \end{aligned}$$

The listener is given to hear $$f_{\text{beat}} = 10\ \text{Hz}$$, so we set

$$10 = f\,\dfrac{2u_0}{v}.$$

Now we substitute $$f = 660\ \text{Hz}$$ and $$v = 330\ \text{m s}^{-1}$$:

$$10 = 660 \times \dfrac{2u_0}{330}.$$

Simplifying the right-hand side, we observe that $$\dfrac{660}{330} = 2$$, so

$$10 = 2 \times 2 u_0 = 4u_0.$$

Solving for $$u_0$$ gives

$$u_0 = \dfrac{10}{4} = 2.5\ \text{m s}^{-1}.$$

Hence, the correct answer is Option B.