Question

Police radar determines the speed of motor vehicles using the same Doppler-shift technique employed for ultrasound in medical diagnostics. Beats are produced by mixing the double Doppler-shifted echo with the original frequency. If $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }\mathbf{109}\mathbf{-}\mathbf{Hz}$microwaves are used and a beat frequency of $\mathbf{150}\mathbf{}\mathbf{Hz}$is produced, what is the speed of the vehicle? (Assume the same Doppler-shift formulas are valid with the speed of sound replaced by the speed of light.)

Short answer

The velocity is$3\mathrm{m}/\mathrm{s}$ .

Step-by-step solution

Define frequency

The number of complete cycles of waves passing a spot in unit time is defined as the frequency.

Evaluate velocity

The apparent frequency and the original frequency are known to be connected via the Doppler effect, which is given by

${\text{f=f}}_0\frac{\left(\mathrm{v}+{\mathrm{v}}_0\right)}{v}$

Where ${\mathrm{f}}_0$is original frequency,$\mathrm{f}$is apparent frequency, $\mathrm{v}$is the velocity of light, and ${\mathrm{v}}_0$is the velocity of the observer.

Also, calculate the apparent frequency by multiplying the original frequency by the beat frequency.

$$\begin{gathered} \mathrm{f}={\mathrm{f}}_0+150\mathrm{Hz} \\ =1.5\times {10}^9\mathrm{Hz}+150\mathrm{Hz} \\ \approx 1.5\times {10}^9\mathrm{Hz} \end{gathered}$$

Substitute the values in the above equation,

$$\begin{gathered} 1.5\times {10}^9\mathrm{Hz}=1.5\times {10}^9\mathrm{Hz}\left(\frac{3\times {10}^8\mathrm{m}/\mathrm{s}+{\mathrm{v}}_0}{3\times {10}^8\mathrm{m}/\mathrm{s}}\right) \\ \frac{1.5\times {10}^9\mathrm{Hz}}{1.5\times {10}^9\mathrm{Hz}}\approx \frac{3\times {10}^8\mathrm{m}/\mathrm{s}+{\mathrm{v}}_0}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ {\mathrm{v}}_0\approx 3\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} 1.5\times {10}^9\mathrm{Hz}=1.5\times {10}^9\mathrm{Hz}\left(\frac{3\times {10}^8\mathrm{m}/\mathrm{s}+{\mathrm{v}}_0}{3\times {10}^8\mathrm{m}/\mathrm{s}}\right) \\ \frac{1.5\times {10}^9\mathrm{Hz}}{1.5\times {10}^9\mathrm{Hz}}\approx \frac{3\times {10}^8\mathrm{m}/\mathrm{s}+{\mathrm{v}}_0}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ {\mathrm{v}}_{0\approx 3\mathrm{m}/\mathrm{s}} \end{gathered}$$

Therefore, the velocity is $3\mathrm{m}/\mathrm{s}$.