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Chapter 22: Problem 59

A police car's radar gun emits microwaves with a frequency of $f_{1}=36.0 \mathrm{GHz}$. The beam reflects from a speeding car, which is moving away at \(43.0 \mathrm{m} / \mathrm{s}\) with respect to the police car. The frequency of the reflected microwave as observed by the police is \(f_{2} .\) (a) Which is larger, \(f_{1}\) or \(f_{2} ?\) (b) Find the frequency difference \(f_{2}-f_{1}\) [Hint: There are two Doppler shifts. First think of the police as source and the speeder as observer. The speeding car "retransmits" a reflected wave at the same frequency at which it observes the incident wave. \(]\)

Short Answer

Expert verified
Answer: \(f_1\) is larger than \(f_2\), and the frequency difference is approximately \(-0.00041751\,\mathrm{GHz}\).

Step by step solution

01

Find the frequency observed by the speeding car

. The frequency observed by the speeding car, given its velocity (43.0 m/s) away from the police, can be found using the Doppler effect formula\\ $$ f_\text{obs} = f_\text{src} \biggl(\frac{c}{c + v_\text{rel}} \biggr) $$ where \(f_\text{obs}\) is the observed frequency (speeding car), \(f_\text{src}\) is the source frequency (police), \(c\) is the speed of light, and \(v_\text{rel}\) is the relative velocity between the source and the observer. Using the given values, we can calculate \(f_\text{obs}\): $$ f_\text{obs} = 36.0\,\mathrm{GHz} \biggl(\frac{3 \times 10^8\,\mathrm{m/s}}{3 \times 10^8\,\mathrm{m/s} + 43.0\,\mathrm{m/s}}\biggr) $$
02

Calculate the frequency observed by the speeding car

. Now, plug in the values into the formula and calculate \(f_\text{obs}\): $$ f_\text{obs} ≈ 35.99979153\,\mathrm{GHz} $$
03

Find the frequency observed by the police car

. Now consider the speeding car as the new source of radiation, and the police car as the observer. We can find the frequency observed by the police car \(f_2\) using the same Doppler effect formula: $$ f_2 = f_\text{obs} \biggl(\frac{c}{c - v_\text{rel}} \biggr) $$ Using the values from step 2, we can calculate \(f_2\): $$ f_2 = 35.99979153\,\mathrm{GHz} \biggl(\frac{3 \times 10^8\,\mathrm{m/s}}{3 \times 10^8\,\mathrm{m/s} - 43.0\,\mathrm{m/s}}\biggr) $$
04

Calculate the frequency observed by the police car

. Now, plug in the values into the formula and calculate \(f_2\): $$ f_2 ≈ 35.99958249\,\mathrm{GHz} $$
05

Answer part (a) of the problem

. Now we can compare the source frequency \(f_1\) and the observed frequency \(f_2\). We can see that \(f_1 = 36.0\,\mathrm{GHz}\) and \(f_2 ≈ 35.99958249\,\mathrm{GHz}\). Therefore, \(f_1\) is larger than \(f_2\).
06

Answer part (b) of the problem

. To find the frequency difference, we can simply subtract the source frequency \(f_1\) from the observed frequency \(f_2\): $$ f_2 - f_1 ≈ 35.99958249\,\mathrm{GHz} - 36.0\,\mathrm{GHz} ≈ -0.00041751\,\mathrm{GHz} $$ The frequency difference \(f_2 - f_1\) is approximately \(-0.00041751\,\mathrm{GHz}\).

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