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Chapter 16: Problem 66

A \(2.00 \mathrm{MHz}\) 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 72 beats per second are detected. The speed of sound in body tissue is \(1500 \mathrm{~m} / \mathrm{s}\). Calculate the speed of the fetal heart wall at the instant this measurement is made.

Short Answer

Expert verified
The speed of the fetal heart wall at the instant this measurement is made is approximately \(0.036 m/s\).

Step by step solution

01

Analyze the given information

From the exercise, the following data can be identified: \nInitial frequency, \(f_i = 2.00 MHz = 2.00 \times 10^6 Hz\),\nDetected beats, \(f_b = 72 Hz\),\nSpeed of sound in the body, \(v = 1500 m/s\).
02

Calculate the frequency of the reflected wave

The detected beat frequency is the difference between the frequencies of the transmitted wave and the reflected wave. So, let's calculate the frequency of the reflected wave, denoted as \(f_r\), using the following equation: \(f_b = f_i - f_r\). Rearranging the formula for \(f_r\) we get: \(f_r = f_i - f_b = 2.00 \times 10^6 Hz - 72 Hz = 1.999928 \times 10^6 Hz\).
03

Apply the Doppler shift formula

We can now apply the Doppler shift formula for the frequency of the wave reflected from the moving heart wall: \(f_r = f_i \times \frac{v + v_0}{v}\), where \(v_0\) is the speed of the heart wall. By rearranging this equation, the speed of the heart wall can be calculated as follows: \(v_0 = \frac{f_r}{f_i} \times v - v.\)
04

Substitute and Solve

Substitute the calculated and given values into the formula: \(v_0 = \frac{1.999928 \times 10^6 Hz}{2.00 \times 10^6 Hz} \times 1500 m/s - 1500 m/s = 0.036 m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acoustics
Acoustics is the science of sound and is crucial in understanding how sound waves behave under various conditions. In the context of ultrasonography, acoustics plays a fundamental role in medical diagnostics. Sound waves, like those used in ultrasonography, have distinct properties such as frequency, wavelength, and speed. These properties help in forming clear images within the body.
The sound waves used in medical ultrasonography are typically high-frequency, ranging from 1 to 20 MHz. Such frequencies are beyond human hearing and provide the resolution required to view minute structures such as a fetal heart. In our exercise, a 2.00 MHz frequency was used, which is sufficient for detailed imaging inside the mother's body.
Key acoustic principles such as the speed of sound, which in bodily tissues is approximately 1500 m/s, aid in calculating crucial details like the Doppler effect involved in our scenario. Understanding these acoustic fundamentals allows for the precise application of ultrasonography in medical settings.
Sound Wave Reflection
Sound wave reflection is a process that occurs when a sound wave hits a surface and bounces back. This principle is used meticulously in ultrasonography. When a sound wave encounters a boundary of a different material or density, such as a fetal heart within a mother's body, part of the wave is reflected back towards the source.
The exercise we examined involves sound waves reflecting off a fetal heart, which is used to gather information about its movement. The frequency difference between the transmitted and reflected wave creates a detectable pattern called the beat frequency, which was 72 beats per second in this scenario.
Reflecting waves are instrumental in diagnostics because they provide data based on how fast or slow tissues are moving. By analyzing these reflected waves, doctors can infer the speed of internal body parts, such as the fetal heart wall's motion, using the Doppler effect as a guide.
Fetal Heart Monitoring
Fetal heart monitoring is a crucial aspect of prenatal care. It involves using acoustic principles and sound wave reflection, as discussed earlier. Through techniques such as Doppler ultrasonography, healthcare providers can monitor and evaluate the fetal heart's condition.
In the exercise, the fetal heart wall's movement was gauged by measuring beat frequency, indicating how the heart is functioning. With the Doppler effect, it's possible to determine the speed at which the heart wall is moving, which we calculated to be approximately 0.036 m/s. This measurement can indicate normal or abnormal conditions, informing necessary medical decisions.
Fetal heart monitoring is vital because it provides a window into the baby's health and growth before birth. The non-invasive nature and reliability of this monitoring make it indispensable in modern obstetric care, offering peace of mind and vital information for both healthcare providers and expectant parents.

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Most popular questions from this chapter

(a) By what factor must the sound intensity be increased to raise the sound intensity level by \(13.0 \mathrm{~dB} ?\) (b) Explain why you don't need to know the original sound intensity.

Standing sound waves are produced in a pipe that is \(1.20 \mathrm{~m}\) long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if (a) the pipe is open at both ends and (b) the pipe is closed at the left end and open at the right end.

You blow across the open mouth of an empty test tube and produce the fundamental standing wave in the \(14.0-\mathrm{cm}\) -long air column in the test tube, which acts as a stopped pipe. (a) What is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

A police car is traveling due east at a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) relative to the earth. You are in a convertible following behind the police car. Your car is also moving due east at \(15.0 \mathrm{~m} / \mathrm{s}\) relative to the earth, so the speed of the police car relative to you is zero. The siren of the police car is emitting sound of frequency \(500 \mathrm{~Hz}\). The speed of sound in the still air is \(340 \mathrm{~m} / \mathrm{s}\) (a) What is the speed of the sound waves relative to you? (b) What is the wavelength of the sound waves at your location? (c) What frequency do you detect?

A stationary police car emits a sound of frequency \(1200 \mathrm{~Hz}\) that bounces off a car on the highway and returns with a frequency of \(1250 \mathrm{~Hz}\). The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving toward or away from the police car? (b) What frequency would the police car have received if it had been traveling toward the other car at \(20.0 \mathrm{~m} / \mathrm{s} ?\)
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