/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 BIO Ultrasound in Medicine. A \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 74

BIO Ultrasound in Medicine. A \(2.00-\) 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 beals. 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 is approximately 0.054 m/s.

Step by step solution

01

Understand the Doppler Effect

When the heart wall moves towards the receiver, the frequency of the returning sound waves increases due to the Doppler effect. This increase in frequency is what causes the beats detected, which are the difference in frequency between the transmitted and received sound waves.
02

Determine Beat Frequency

The beat frequency given is 72 Hz. Beat frequency is the absolute difference between the received (reflected) frequency and the transmitted frequency: \( f_{beat} = |f_{r} - f_{t}| \). Given that the transmitted frequency \( f_t \) is 2 MHz, or \(2 \times 10^{6}\) Hz, we will proceed to find the speed of the heart wall.
03

Use the Doppler Shift Formula for Moving Source

For a source moving towards the observer, the received frequency is \( f_r = f_t \left(\frac{v + v_o}{v - v_s}\right) \). Here, \(v\) is the speed of sound (1500 m/s), \(v_o\) is the speed of the observer (zero, since the receiver is stationary), and \(v_s\) is the speed of the heart wall. We need to solve this for \(v_s\).
04

Relate Beat Frequency to Doppler Shift

Since \( f_{beat} = f_r - f_t \), we substitute for \( f_r \) using the Doppler formula: \( 72 = 2 \times 10^6 \left(\frac{1500}{1500 - v_s}\right) - 2 \times 10^6 \). Rearrange this to solve for \( v_s \).
05

Solve the Equation for the Speed of Heart Wall

Simplify the equation: \[72 = 2 \times 10^6 \times \left( \frac{1}{1500 - v_s} \times 1500 \right) - 2 \times 10^6\]This equation can be further simplified and solved for \(v_s\):\[72 = 2 \times 10^6 \times \left(\frac{1}{1500 - v_s}\right) - 1 \]Solving this will give us \(v_s = 0.054 \) m/s.
06

Calculate the Result

Rearrange the equation to isolate \( v_s \):\[72 + 2 \times 10^6 = 2 \times 10^6 \cdot \frac{1500}{1500-v_s} \]Then, solve for \(v_s\) using algebraic manipulation. Simplifying this further gives the speed of the fetal heart wall:\[ v_s = \frac{(72/2\times10^6) \times 1500}{2\times10^6 - 72/2\times10^6}\]Finally, calculate to find the value of \( v_s = 0.054 \) m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ultrasound in Medicine
Ultrasound is a fascinating tool used in medical diagnostics. It involves sound waves with frequencies higher than the human hearing range. In medicine, ultrasound is often used to create images, as it can safely penetrate body tissues. This allows doctors to observe internal structures like the development of a fetus in a woman's uterus.

The key to ultrasound in medicine is its ability to reflect sound waves off different tissues. These reflections create images based on the strength and timing of the returning waves. This non-invasive method is safe for both mother and baby, making it invaluable in prenatal care.
Beat Frequency
In the context of sound waves, a beat frequency occurs when two similar frequencies interfere. This phenomenon is easily heard when tuning musical instruments, but it's also crucial in medical physics.

When sound waves from a transmitter meet those reflected back from a moving surface, like a fetal heart wall, they overlap. The difference in frequency, known as the beat frequency, is detected when these waves are combined. This frequency provides essential information about the velocity of the heart wall. By analyzing it, medical practitioners can gather critical insights into the health and movement of body tissues.
Sound Wave Propagation
Sound wave propagation is the movement of sound waves through a medium. In the case of ultrasound, the medium is body tissue. Sound travels as a wave of pressure variations, transiting through different layers within the body.

When these waves move through tissues, their speed can be influenced by factors like tissue density and composition. To calculate things like the speed of a moving heart wall, understanding the propagation of sound waves is crucial. The medium's properties determine how fast the waves travel, which affects calculations, such as the speed of the fetal heart wall using the Doppler effect.
Medical Physics
Medical physics is a field that combines physics principles with medical insight to innovate and improve healthcare. It includes the application of physics in diagnosis and treatment, ensuring safety and effectiveness.

One prominent example is the use of the Doppler effect in ultrasound. By leveraging changes in frequency due to movement, it allows for real-time analysis of organs and tissues. This helps in monitoring physiological and pathological processes.
  • Ensures precise diagnostic techniques.
  • Facilitates non-invasive procedures.
  • Promotes patient safety and comfort.
Thus, medical physics is pivotal in evolving healthcare technologies and methodologies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Tuning a Violin. A violinist is tuning her instrument to concert \(\mathrm{A}(440 \mathrm{Hz}) .\) She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 3 \(\mathrm{Hz}\) , which increases to 4 \(\mathrm{Hz}\) when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3 -Hz beat? (b) To get her violin perfectly tuned to concert \(\Lambda,\) should she tighten or loosen her string from what it was when she heard the 3 -Hz beat?

The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . (b) If the pipe is filled with air at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

CP The sound from a trumpet radiates uniformly in all directions in \(20^{\circ} \mathrm{C}\) air. At a distance of 5.00 \(\mathrm{m}\) from the trumpet the sound intensity level is 52.0 \(\mathrm{dB}\) . The frequency is 587 \(\mathrm{Hz}\) . (a) What is the pressure amplitude at this distance? (b) What is the displacement amplitude? (c) At what distance is the sound intensity level 30.0 \(\mathrm{dB}\) ?

On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz}\) . If the stationary female hears a tone of 1240 \(\mathrm{Hz}\) , what is the speed of sound in the atmosphere of Arrakis?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electromagnetism

Read Explanation

Medical Physics

Read Explanation

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Electricity and Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.