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

The source is at rest and has frequency $$f_s = 500\ \text{Hz}$$. We recall the Doppler formula for a moving observer when the source is stationary:

$$f' = f_s\left(\dfrac{v \;\pm\; v_o}{v}\right)$$

where $$v = 300\ \text{m s}^{-1}$$ is the speed of sound, $$v_o$$ is the speed of the observer (always taken positive), and the sign in the numerator is

  • “+” if the observer moves towards the source (frequency increases),
  • “−” if the observer moves away from the source (frequency decreases).

One observer hears a reduced frequency $$f'_1 = 480\ \text{Hz}$$, so he must be moving away. Using the “−” sign we write

$$480 = 500\left(\dfrac{v - v_{o1}}{v}\right).$$

Dividing both sides by 500, we obtain

$$\dfrac{480}{500} = \dfrac{v - v_{o1}}{v}.$$

Simplifying the left side:

$$0.96 = 1 - \dfrac{v_{o1}}{v}.$$

Now we isolate $$v_{o1}$$:

$$\dfrac{v_{o1}}{v} = 1 - 0.96 = 0.04,$$ $$v_{o1} = 0.04\,v = 0.04 \times 300 = 12\ \text{m s}^{-1}.$$

The other observer hears an increased frequency $$f'_2 = 530\ \text{Hz}$$, so he is moving towards the source. Using the “+” sign we write

$$530 = 500\left(\dfrac{v + v_{o2}}{v}\right).$$

Dividing both sides by 500 gives

$$\dfrac{530}{500} = \dfrac{v + v_{o2}}{v}.$$

Simplifying the left side:

$$1.06 = 1 + \dfrac{v_{o2}}{v}.$$

Now we isolate $$v_{o2}$$:

$$\dfrac{v_{o2}}{v} = 1.06 - 1 = 0.06,$$ $$v_{o2} = 0.06\,v = 0.06 \times 300 = 18\ \text{m s}^{-1}.$$

Therefore the two observed speeds are $$12\ \text{m s}^{-1}$$ (for the 480 Hz observer) and $$18\ \text{m s}^{-1}$$ (for the 530 Hz observer).

Hence, the correct answer is Option C.