The Doppler Effect and ultrasound waves (f=2.25 MHz) are used to monitor the hea

Question

The Doppler Effect and ultrasound waves (f=2.25 MHz) are used to monitor the heartbeat of a human fetus inside the motor. The waves are directed at the surface of the beating heart, which movies alternately toward and away from the wave source. This cause Doppler shifts to both increase and decrease the original frequency. The original wave and Doppler-shifted waves are combined and the beat frequency they create is measured. From this information, the velocity of the surface of the fetal heart forward and back can be determined. Assume the speed of sound in the human body is c = 1.46 km/s and the beat frequency is measured to be f beat =260Hz. Determine the velocity of the surface of the fetal heart.

Explanation / Answer

SOLUTION:

A pulsation occurs when two waves of similar frequency meet at the same point. In this case the 260 Hz pulsation is the difference between the received wave frequency f'' from the heart's surface and the original frequency

fs = 2.25x106 Hz, f''- fs = 260 Hz

To calculate the received wave frequency f'' we must realize that there are two Doppler shifts that generate f'':
The heart, moving with velocity v0 "detects" a wave with a frequency f' slightly greater than fs given by :

f' = [(v+v0)/v]fs
The heart "reflects" a new sound of frequency:

f'' = [(v/(v-v0)]f'.

Replacing in f''-fs = 260 Hz results:

f''-fs = [(v/(v-v0)][(v+v0)/v]fs - fs = 260,

where v0 is the asked heart's velocity, v is the sound velocity in the tissues and fs is the ultrasonic source frequency.

Replacing values and solving for v0 it is finally obtained v0.

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