/*! 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 1 Diagnostic ultrasound of frequen... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Diagnostic ultrasound of frequency \(3.80 \mathrm{MHz}\) is used to examine tumors in soft tissue. (a) What is the wavelength in air of such a sound wave? (b) If the speed of sound in tissue is \(1500 \mathrm{~m} / \mathrm{s}\), what is the wavelength of this wave in tissue?

Short Answer

Expert verified
(a) 0.0903 mm in air; (b) 0.395 mm in tissue.

Step by step solution

01

Understanding the Problem

We need to find the wavelengths of ultrasound waves with a frequency of \(3.80 \text{ MHz}\) in two different media: air and soft tissue. The speed of sound in air and tissue is given or can be known.
02

Use the Formula for Wavelength

The wavelength \(\lambda\) of a wave can be calculated using the formula: \(\lambda = \frac{v}{f}\), where \(v\) is the speed of sound and \(f\) is the frequency of the wave.
03

Calculate the Wavelength in Air

Assuming the speed of sound in air is approximately \(343 \text{ m/s}\), the wavelength in air is calculated as:\[ \lambda_{\text{air}} = \frac{343 \text{ m/s}}{3.80 \times 10^6 \text{ Hz}} = 9.03 \times 10^{-5} \text{ meters} \] or \(0.0903 \text{ mm}\).
04

Calculate the Wavelength in Tissue

With the speed of sound in tissue as \(1500 \text{ m/s}\), the wavelength in tissue is calculated as:\[ \lambda_{\text{tissue}} = \frac{1500 \text{ m/s}}{3.80 \times 10^6 \text{ Hz}} = 3.95 \times 10^{-4} \text{ meters} \] or \(0.395 \text{ mm}\).

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 91Ó°ÊÓ!

Key Concepts

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

Wavelength Calculation
Understanding how to calculate wavelength is important in ultrasound physics. Wavelength refers to the distance between two consecutive points in phase on a wave, most commonly peak to peak. To find the wavelength, you use the formula:
  • \( \lambda = \frac{v}{f} \)
where
  • \( \lambda \) is the wavelength,
  • \( v \) is the speed of sound in the medium, and
  • \( f \) is the frequency of the wave.
In our exercise, we calculate the wavelength for an ultrasound wave with a frequency of 3.80 MHz in different media. In air, where the speed of sound is around 343 m/s, the wavelength is calculated to be approximately 0.0903 mm. This is done by substituting the values into the formula:\[ \lambda_{\text{air}} = \frac{343 \text{ m/s}}{3.80 \times 10^6 \text{ Hz}} = 0.0903 \text{ mm} \]In soft tissue, where the speed of sound is 1500 m/s, the wavelength becomes:\[ \lambda_{\text{tissue}} = \frac{1500 \text{ m/s}}{3.80 \times 10^6 \text{ Hz}} = 0.395 \text{ mm} \]This shows how changing the medium affects the wavelength, even if the frequency remains the same.
Sound Propagation
Sound waves travel differently depending on the medium they propagate through. This is because the speed of sound varies in different materials, affecting how the wave moves and its subsequent properties like wavelength. In general, sound waves travel faster in denser media.
For instance:
  • Speed of sound in air: Approximately 343 m/s.
  • Speed of sound in soft tissue: Approximately 1500 m/s.
This disparity in speed affects diagnostic imaging using ultrasounds. As the sound wave travels from one medium to another, its speed changes, impacting the wavelength.
Understanding this concept is essential for interpreting ultrasound imaging, as different body tissues will influence sound speed and affect the clarity and accuracy of the images produced. Adjusting the frequency or utilizing specific equipment settings can help mitigate these effects and provide clearer medical images.
Frequency in Medical Imaging
In medical imaging, particularly ultrasound, frequency plays a critical role in the quality and penetration of the images produced. Frequency is defined as the number of wave cycles that occur in one second and is measured in Hertz (Hz).
  • Higher frequencies provide better resolution but do not penetrate deeply into tissue.
  • Lower frequencies penetrate deeper but offer lower resolution.
The frequency used in medical ultrasound typically ranges from 2 MHz to 15 MHz. Our exercise involves a frequency of 3.80 MHz, a common setting for examining soft tissue structures like tumors. Choosing the appropriate frequency ensures that the ultrasound can effectively visualize the area of interest.
Understanding frequency allows healthcare professionals to customize imaging techniques to achieve optimal balance between image resolution and depth penetration, crucial for accurate diagnosis and patient care. This customization can improve the ultrasound's capability to distinguish fine details in soft tissues, influencing the diagnosis and treatment plans.

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

An acoustic burglar alarm consists of a source emitting waves of frequency \(30.0 \mathrm{kHz}\). What is the beat frequency between the source waves and the waves reflected from an intruder walking at an average speed of \(0.950 \mathrm{~m} / \mathrm{s}\) directly away from the alarm?

A tuning fork of unknown frequency makes \(4.00\) beats per second with a standard fork of frequency \(384 \mathrm{~Hz}\). The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

Earthquakes generate sound waves inside Earth. Unlike a gas, Earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of \(\mathrm{S}\) waves is about \(4.5 \mathrm{~km} / \mathrm{s}\), and that of \(\mathrm{P}\) waves \(8.0 \mathrm{~km} / \mathrm{s}\). A seismograph records \(\mathrm{P}\) and \(\mathrm{S}\) waves from an earthquake. The first \(\mathrm{P}\) waves arrive \(3.5 \mathrm{~min}\) before the first \(\mathrm{S}\) waves. If the waves travel in a straight line, how far away did the earthquake occur?

You have five tuning forks that oscillate at close but different resonant frequencies. What are the (a) maximum and (b) minimum number of different beat frequencies you can produce by sounding the forks two at a time, depending on how the resonant frequencies differ?

A stationary detector measures the frequency of a sound source that first moves at constant velocity directly toward the detector and then (after passing the detector) directly away from it. The emitted frequency is \(f\). During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\). If \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=\) \(0.200\), what is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

See all solutions

Recommended explanations on Physics Textbooks

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.