You are driving along a highway at \(30.0 \mathrm{~m} / \mathrm{s}\) when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is \(1300 \mathrm{~Hz}\). Right after the police car passes you, the frequency of the siren that you hear is \(1280 \mathrm{~Hz}\). a) How fast was the police car moving? b) You are so nervous after the police car passes you that you pull off the road and stop. Then you hear another siren, this time from an ambulance approaching from behind. The frequency of its siren that you hear is \(1400 \mathrm{~Hz}\). Once it passes, the frequency is \(1200 \mathrm{~Hz}\). What is the actual frequency of the ambulance's siren?

Short Answer

Expert verified
Based on the given scenario, the speed of the police car is 40.0 m/s, and the actual frequency of the ambulance's siren is 1260 Hz.

Step by step solution

01

Doppler effect when the siren is approaching

When the police car is approaching with frequency \(f_{approaching}=1300 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{approaching} = f \frac{v_{sound} + v_{listener}}{v_{sound} - v_{source}}$$ Where \(v_{listener} = 30.0 \mathrm{~m} / \mathrm{s}\) and \(v_{source}\) is the speed of the police car.
02

Doppler effect when the siren is receding

When the police car is moving away with frequency \(f_{receding} = 1280 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{receding} = f \frac{v_{sound} - v_{listener}}{v_{sound} + v_{source}}$$
03

Find the speed of the police car

Divide the equations obtained in Step 1 and Step 2: $$\frac{f_{approaching}}{f_{receding}} = \frac{v_{sound} + v_{listener}}{v_{sound} - v_{source}} \times \frac{v_{sound} + v_{source}}{v_{sound} - v_{listener}}$$ Substitute the known values and solve for \(v_{source}\): $$\frac{1300}{1280} = \frac{370}{v_{sound} - v_{source}} \times \frac{v_{sound} + v_{source}}{310}$$ Simplify and solve for \(v_{source}\), we get: $$v_{source} = 40.0 \mathrm{~m} / \mathrm{s}$$ So, the speed of the police car is \(40.0 \mathrm{~m} / \mathrm{s}\). **Part b**
04

Doppler effect when the siren is approaching

When the ambulance is approaching with frequency \(f_{approaching}=1400 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{approaching} = f_0 \frac{v_{sound}}{v_{sound} - v_{source_a}}$$ Where \(v_{source_a}\) is the speed of the ambulance.
05

Doppler effect when the siren is receding

When the ambulance is moving away with frequency \(f_{receding} = 1200 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{receding} = f_0 \frac{v_{sound}}{v_{sound} + v_{source_a}}$$ Where \(f_0\) is the actual frequency of the ambulance's siren.
06

Find the actual frequency of the ambulance's siren

Divide the equations obtained in Step 1 and Step 2: $$\frac{f_{approaching}}{f_{receding}} = \frac{v_{sound} - v_{source_a}}{v_{sound} + v_{source_a}}$$ Substitute the known values and solve for \(v_{source_a}\): $$\frac{1400}{1200} = \frac{340 - v_{source_a}}{340 + v_{source_a}}$$ Solving for \(v_{source_a}\), we get: $$v_{source_a} = 25.0 \mathrm{~m} / \mathrm{s}$$ Now substitute the value of \(v_{source_a}\) in either Step 1 or Step 2 formula and solve for \(f_0\). Substituting in Step 1 formula, we get: $$f_0 = \frac{f_{approaching}(v_{sound}-v_{source_a})}{v_{sound}}$$ $$f_0 = \frac{1400 \times (340 - 25)}{340} = 1260 \mathrm{~Hz}$$ So, the actual frequency of the ambulance's siren is \(1260 \mathrm{~Hz}\).

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Most popular questions from this chapter

Two people are talking at a distance of \(3.0 \mathrm{~m}\) from where you are, and you measure the sound intensity as \(1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). Another student is \(4.0 \mathrm{~m}\) away from the talkers. What sound intensity does the other student measure?

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