Question

(III) The Doppler effect using ultrasonic waves of frequency $2.25\times {10}^6\mathrm{H}\mathrm{z}$ is used to monitor the heartbeat of a fetus. A (maximum) beat frequency of 240 Hz is observed. Assuming that the speed of sound in tissue is 1540 m/s, calculate the maximum velocity of the surface of the beating heart.

Short answer

The maximum velocity of the surface is $0.082\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given Data

The frequency of the wave is$f=2.25\times {10}^6\mathrm{H}\mathrm{z}$.

The beat frequency is$f_{\mathrm{b}}=240\mathrm{H}\mathrm{z}$.

The speed of the sound in tissue is $v_{\mathrm{s}}=1540\mathrm{m}/\mathrm{s}$.

Understanding the Doppler shifts

In this problem, there are two Doppler shifts, the first one occurs when the heart receives the original signal and the other occurs for the detector reflected signal.

Evaluating the relation between beat frequency and original frequency

The relation from Doppler shift formula is given by,

$f_0=f\left(\frac{v_{\mathrm{s}}-v}{v_{\mathrm{s}}+v}\right)$

Here, $v_{\mathrm{s}}$ is the speed of the sound, $f_0$ is the frequency for the detector and v is the velocity of the blood.

The relation of beat frequency is given by,

$\begin{array}{rl}c{f}_{\mathrm{b}}& =f-f_0\\ f_{\mathrm{b}}& =\left(f-f\left(\frac{v_{\mathrm{s}}-v}{v_{\mathrm{s}}+v}\right)\right)\\ f_{\mathrm{b}}& =f\left(\frac{\left(v_{\mathrm{s}}+v\right)-\left(v_{\mathrm{s}}-v\right)}{v_{\mathrm{s}}+v}\right)\\ f_{\mathrm{b}}& =f\left(\frac{2v}{v_{\mathrm{s}}+v}\right)\\ v& =v_{\mathrm{s}}\left(\frac{f_{\mathrm{b}}}{2f-f_{\mathrm{b}}}\right)\end{array}$

Evaluating the maximum velocity

On plugging the values in the above relation.

$\begin{array}{rl}lv& =\left(1540\mathrm{m}/\mathrm{s}\right)\left(\frac{240\mathrm{H}\mathrm{z}}{2\left(2.25\times {10}^6\mathrm{H}\mathrm{z}\right)-240\mathrm{H}\mathrm{z}}\right)\\ v& =0.082\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=0.082\mathrm{m}/\mathrm{s}$ is the required maximum velocity.