Question

A police car's radar gun emits microwaves with a frequency of $f_{1}=7.50 \mathrm{GHz}$. The beam reflects from a speeding car, which is moving toward the police car at \(48.0 \mathrm{m} / \mathrm{s}\) with respect to the police car. The speeder's radar detector detects the microwave at a frequency \(f_{2}\). (a) Which is larger, \(f_{1}\) or \(f_{2} ?\) (b) Find the frequency difference \(f_{2}-f_{1}\).

Short answer

Answer: The detected frequency (f_2) is larger than the emitted frequency (f_1), and the frequency difference (f_2 - f_1) is approximately 1.44 × 10^6 Hz.

Step-by-step solution

Identify given values and find initial frequency

We are given the frequency of the emitted microwaves, \(f_1 = 7.50 \rm{GHz}\), and the speed of the car, \(v_{car} = 48.0 \rm{m/s}\). We are also given that the speed of light is \(c = 2.998 \times 10^8 \mathrm{m/s}\).

Calculate the frequency of the detected microwaves, \(f_2\)

Using the formula for the Doppler effect, we can find the detected frequency \(f_2\): $$f_2 = f_1 \left( \frac{c + v_{car}}{c} \right)$$ Substituting the given values, we get: $$f_2 = 7.50 \times 10^9 \left( \frac{2.998 \times 10^8 + 48.0}{2.998 \times 10^8} \right)$$

Solve for \(f_2\)

Calculate the detected frequency \(f_2\): $$f_2 = 7.50 \times 10^9 \left( \frac{2.998 \times 10^8 + 48.0}{2.998 \times 10^8} \right) \approx 7.5014 \times 10^9 \mathrm{Hz}$$

Compare \(f_1\) and \(f_2\)

Since \(f_2 = 7.5014 \times 10^9 \gt 7.50 \times 10^9 = f_1\), we can conclude that the detected frequency, \(f_2\), is larger than the emitted frequency, \(f_1\).

Find the frequency difference \(f_2 - f_1\)

Calculate the frequency difference using the values of \(f_1\) and \(f_2\). $$f_2 - f_1 = (7.5014 \times 10^9) - (7.50 \times 10^9) \approx 1.44 \times 10^6 \mathrm{Hz}$$ The frequency difference, \(f_2 - f_1\), is approximately \(1.44 \times 10^6 \mathrm{Hz}\).