Question

An ambulance traveling at \(44 \mathrm{m} / \mathrm{s}\) approaches a car heading in the same direction at a speed of \(28 \mathrm{m} / \mathrm{s}\). The ambulance driver has a siren sounding at \(550 \mathrm{Hz}\). At what frequency does the driver of the car hear the siren?

Short answer

Answer: The driver of the car hears the ambulance siren at a frequency of approximately 579.1 Hz.

Step-by-step solution

Identify the given variables

We are given the following variables: - Speed of the ambulance, \(v_A = 44 \, \text{m/s}\) - Speed of the car, \(v_C = 28 \, \text{m/s}\) - Frequency of the siren, \(f_s = 550 \, \text{Hz}\) We need to find the frequency of the sound received by the car driver.

Calculate the relative speed

Both the ambulance and the car are moving in the same direction, so their relative speed will be the difference in their speeds. Relative speed, \(v_R = v_A - v_C = 44 \, \text{m/s} - 28 \, \text{m/s} = 16 \, \text{m/s}\)

Apply the Doppler effect equation

The Doppler effect equation for a moving source and moving listener, both in the same direction and moving away from each other, is given by: $$f_R = \frac{f_s(v - v_L)}{v - v_S}$$ Where \(f_R\) is the received frequency by the listener (car driver), \(f_s\) is the source frequency (siren frequency), \(v\) is the speed of sound in air (approximately \(343 \, \text{m/s}\)), \(v_L\) is the listener's (car) speed, and \(v_S\) is the source's (ambulance) speed. We have \(f_s = 550 \, \text{Hz}\), \(v = 343 \, \text{m/s}\), \(v_L = v_C = 28 \, \text{m/s}\), and \(v_s = v_A = 44 \, \text{m/s}\). Plug in these values into the Doppler effect equation: $$f_R = \frac{550 \times (343 - 28)}{343 - 44}$$

Calculate the received frequency

Calculate the received frequency: $$f_R = \frac{550 \times (315)}{299} = \frac{173250}{299} \approx 579.1 \, \text{Hz}$$ The driver of the car hears the ambulance siren at a frequency of approximately \(579.1 \, \text{Hz}\).