Question: Figure 17-50 shows a transmitter and receiver of waves contained in a single instrument. It is used to measure the speed uof a target object (idealized as a flat plate) that is moving directly toward the unit, by analyzing the waves reflected from the target. What is uif the emitted frequency is 18.0 kHzand the detected frequency (of the returning waves) is 22.2 kHz?

• i) The frequency emitted by the transmitter is f =18.0kHz or$1.8\times {10}^{4}Hz$
• ii)The frequency detected by the detector is$f\text{'}=22.2kHzor2.22\times {10}^{4}Hz$

Apply Doppler Effect for approaching and reflecting waves and get two equations. By solving them, we can get the speed of the target object.

Formulae:

The frequency received by the observer or the source according to Doppler’s Effect, (when detector at rest and source is approaching detector)

$f\text{'}=f\times \frac{V}{V-V_S}$ (i)

(The source at rest and detector is approaching source)

$f\text{'}=f\times \frac{V+V_D}{V}$ (ii)

Let us assume that the target isthedetector. We have the equation (ii) of Doppler Effect for stationary source and moving detector whenthedetector moves toward the source given as:

$f_1=1.8\times {10}^4\times \frac{343+u}{343}$ …… (a)

Now, we consider the reflected wave fromthetarget asthesource wave.

We have the equation (i) of Doppler Effect for stationary detector and moving source whenthesource is moving towardsthedetector as:

Substitute the values in equation (a) and solve:

role="math" $$\begin{gathered} 2.22\times {10}^4=f_1\times \frac{343}{343-u} \\ 2.22\times {10}^4=1.8\times {10}^4\times \frac{343+u}{343}\times \frac{343}{343-u} \\ \frac{2.22\times {10}^4}{1.8\times {10}^4}\times \left(343-u\right)=\left(343+u\right) \\ 423.033-1.233\times u=343+u \\ Solvinffurtheras: \\ 423.033-343=2.233\times u \\ \frac{80.033}{2.233}=u \\ u=35.84\frac{m}{s} \end{gathered}$$

Solve further as:

Hence, the value of the speed of the target object is$35.84\frac{m}{s}$