Question

A 2.15-MHz sound wave travels through a pregnant woman’s abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 90 beats per second are detected. The speed of sound in body tissue is 1510 m/s. Calculate the speed of the fetal heart wall at the instant this measurement is made.

Answer 1

Answer:

v= 0.0316 m/s

Explanation:

We need to use the Doppler Effect defined as the change in frequency of a wave in relation to an observer who is moving relative to the wave source.

Notation

Let v= magnitude of the heart wall speed

V= speed  of sound

fh= the frequency the heart receives (and  reflects)

fi= original frequency

ff= reflected frequency

fb= frequency for the beats

Apply the Doppler Effect formula

Since the heart is moving observer then the device is a stationary source, and we have this formula

fh = [(V+ v)/(v)] fi  (1)

We can consider the heart as moving source and the device as a stationary observer, and we have this formula

ff = [(V)/(V-v)] fh  (2)

The frequency for the beats would be the difference from the original and the reflected frequency

fb = ff -fi (3)

Replacing equations (1) and (2) into equation (3) we have:

$f_b = \frac{V}{V-v} \frac{V+v}{V}f_i - f_i$

$f_b = f_i ( \frac{V+v}{V-v} -1)$

fb = fi(V+v -V+v)/(V-v)

$f_b = \frac{2v}{V-v}$

Solving for v we have:

$v = V (\frac{f_b}{2f_o - f_b})$

$v = 1510 m/s (\frac{90 Hz}{2∗2150000Hz - 90 Hz})= 0.0316 m/s$