Question

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 \(1400 .\) Hz. Once it passes, the frequency is \(1200 .\) Hz. What is the actual frequency of the ambulance's siren?

Short answer

Answer: The police car was moving at 33.6 m/s, and the actual frequency of the ambulance's siren was 1279.34 Hz.

Step-by-step solution

Write down the Doppler effect formula for moving observer and moving source

The Doppler effect formula for a moving observer and a moving source is given by:$$f = \frac{f_0}{1-\frac{v_{\text{observer}}}{v_s}\pm \frac{v_{\text{source}}}{v_s}}.$$

Write equations for the police car passing the observer

As the police car approaches, let's use the positive sign in the Doppler effect formula: $$f_{approach} = \frac{f_0}{1-\frac{v_{\text{observer}}}{v_s} + \frac{v_{\text{source}}}{v_s}}.$$ After the police car passes the observer, we will use the negative sign: $$f_{recede} = \frac{f_0}{1-\frac{v_{\text{observer}}}{v_s} - \frac{v_{\text{source}}}{v_s}}.$$

Substitute known values and solve for the velocity of the police car

Given that \(f_{approach} = 1300 \mathrm{~Hz}\), \(f_{recede} = 1280 \mathrm{~Hz}\), and \(v_{\text{observer}} = 30.0 \mathrm{~m} / \mathrm{s}\), we can write the equations: $$1300 = \frac{f_0}{1-\frac{30}{343} + \frac{v_{\text{source}}}{343}},$$ $$1280 = \frac{f_0}{1-\frac{30}{343} - \frac{v_{\text{source}}}{343}}.$$ Now, solve the equations simultaneously to find the actual frequency \(f_0\) and the source velocity \(v_{\text{source}}\) (police car's velocity): $$f_0 = 1300 \left(1-\frac{30}{343} + \frac{v_{\text{source}}}{343}\right) = 1280 \left(1-\frac{30}{343} - \frac{v_{\text{source}}}{343}\right).$$ After solving for \(v_{\text{source}}\), we get: \(v_{\text{source}} = 33.6 \mathrm{~m} / \mathrm{s}\). Answer to part (a): The police car was moving at \(33.6 \mathrm{~m} / \mathrm{s}\).

Write down the equations for the ambulance siren

We will use the formula for the Doppler effect as the ambulance approaches and recedes from the observer at rest, since they have now stopped: $$f_{approach\_ambulance} = \frac{f_0}{1 + \frac{v_{\text{source\_ambulance}}}{v_s}},$$ $$f_{recede\_ambulance} = \frac{f_0}{1 - \frac{v_{\text{source\_ambulance}}}{v_s}}.$$

Substitute known values and solve for the actual frequency of the ambulance's siren

Given that \(f_{approach\_ambulance} = 1400 \mathrm{~Hz}\), \(f_{recede\_ambulance} = 1200 \mathrm{~Hz}\) and we assume both frequencies are related to the same actual frequency, we can write the equations: $$1400 = \frac{f_0}{1+\frac{v_{\text{source\_ambulance}}}{v_s}},$$ $$1200 = \frac{f_0}{1-\frac{v_{\text{source\_ambulance}}}{v_s}}.$$ We can divide the two equations to get the velocity of the ambulance: $$\frac{1400}{1200} = \frac{1+\frac{v_{\text{source\_ambulance}}}{v_s}}{1-\frac{v_{\text{source\_ambulance}}}{v_s}}.$$ Solving for \(v_{\text{source\_ambulance}}\), we find that \(v_{\text{source\_ambulance}} = 16.86 \mathrm{~m} / \mathrm{s}.\) Now we can substitute \(v_{\text{source\_ambulance}}\) value into either of the equations and solve for \(f_0\). We choose the approach equation: $$1400 = \frac{f_0}{1+\frac{16.86}{343}}.$$ Solving for \(f_0\), we get \(f_0 = 1279.34 \mathrm{~Hz}\). Answer to part (b): The actual frequency of the ambulance's siren is \(1279.34 \mathrm{~Hz}\).