/*! 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 4 Ultrasound Imaging. Sound having... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 4

Ultrasound Imaging. Sound having frequencies above the range of human hearing (about \(20,000 \mathrm{Hz}\) ) is called ultrasound. Waves above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel through body tissue with a speed of 1500 \(\mathrm{m} / \mathrm{s} .\) For a good, detailed image, the wavelength should be no more than 1.0 \(\mathrm{mm} .\) What frequency sound is required for a good scan?

Short Answer

Expert verified
The required frequency for a good ultrasound scan is 1,500,000 Hz (or 1.5 MHz).

Step by step solution

01

Understand the Relationship between Speed, Frequency, and Wavelength

The relationship between speed, frequency, and wavelength of a wave is given by the formula \( v = f \times \lambda \), where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) is the wavelength.
02

Convert the Wavelength to Meters

Given that the maximum wavelength for a detailed image is 1.0 mm, convert this to meters: \( 1.0 \text{ mm} = 0.001 \text{ m} \). This step is important because the speed of sound is given in meters per second.
03

Substitute Values into the Wave Equation

Using the equation \( v = f \times \lambda \), substitute the given wave speed \( v = 1500 \text{ m/s} \) and the converted wavelength \( \lambda = 0.001 \text{ m} \) into the equation to solve for frequency \( f \):\[ 1500 = f \times 0.001 \]
04

Solve for Frequency

Rearrange the equation to solve for the frequency: \( f = \frac{1500}{0.001} \). Now calculate:\[ f = 1500000 \text{ Hz} \]
05

Confirm the Frequency is Above 20,000 Hz

Check that the calculated frequency \( f = 1500000 \text{ Hz} \) is indeed above the human hearing range of 20,000 Hz. Since it is above, this frequency is classified as ultrasound and is appropriate for imaging within the specified requirements.

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.

Wave Speed
Wave speed is a crucial concept in understanding ultrasound imaging. It refers to how fast a wave travels through a particular medium. In the context of ultrasound, sound waves travel through the body at a speed determined by the properties of the tissues. For medical imaging, we use the fact that ultrasound waves travel at approximately 1500 meters per second through body tissues.
This speed is critical because it determines how effectively sound waves can penetrate tissues and return useful images.
  • Wave speed affects how sharp and detailed the ultrasound image is.
  • It combines with frequency and wavelength to form a complete picture of wave behavior.
Understanding wave speed helps technicians adjust settings to optimize image quality in ultrasound diagnostics.
Frequency Calculation
Frequency calculation in ultrasound imaging helps determine how detailed the images will be. The frequency of a wave is how many cycles it completes in a second. In ultrasound, sound waves' frequency must be above 20,000 Hz, as these frequencies penetrate the body to provide clear images.
The frequency is calculated using the formula relating wave speed and wavelength:\[ f = \frac{v}{\lambda} \]
where:
  • \( v \) is the wave speed (1500 m/s in tissue).
  • \( \lambda \) is the wavelength (expressed in meters).
For a wavelength of 0.001 meters, the frequency necessary for a good scan is 1,500,000 Hz. This high frequency ensures that the waves can reflect off small structures, improving image resolution.
Wavelength Conversion
Wavelength is the distance between consecutive peaks of a wave. For ultrasound imaging, ensuring the correct wavelength is critical to acquiring clear images. Often, specifications use millimeters, but calculations typically use meters. Converting from millimeters to meters is straightforward and necessary:
  • 1 millimeter = 0.001 meters
This conversion helps fit into the equation \( v = f \times \lambda \), where calculations are consistent in units. A typical scenario requires the wavelength to be no more than 1.0 mm (or 0.001 meters), ensuring that the ultrasound frequency is high enough to resolve fine details and provide clear, accurate images.
Acoustic Waves
Acoustic waves are fundamental to ultrasound imaging. These are sound waves that can travel through different media, like human tissues. For imaging, ultrasound uses frequencies higher than human hearing can detect, referred to as ultrasonic acoustic waves.
These acoustic waves interact with tissue by:
  • Travelling through, being absorbed, or scattered by tissues.
  • Reflecting back to a probe, which translates these reflections into images.
The quality of these images depends greatly on how well these waves penetrate tissues and the frequency and wavelength chosen. Understanding these interactions is critical for technicians and medical professionals to optimize and interpret ultrasound scans correctly.

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

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 \(\mathrm{Hz}\) , amplitude 5.00 \(\mathrm{mm}\) , and wavelength 0.600 \(\mathrm{m} .\) (a) How long does it take the wave to travel a distance of 8.00 \(\mathrm{m}\) along the length of the string? (b) How long does it take a point on the string to travel a distance of \(8.00 \mathrm{m},\) once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?

A light wire is tightly stretched with tension \(F .\) Trans- verse traveling waves of amplitude \(A\) and wavelength \(\lambda_{1}\) carry average power \(P_{\mathrm{av}, 1}=0.400 \mathrm{W}\) . If the wavelength of the waves is doubled, so \(\lambda_{2}=2 \lambda_{1},\) while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{\mathrm{av}, 2}\) carried by the waves?

One string of a certain musical instrument is 75.0 \(\mathrm{cm}\) long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 \(\mathrm{m} / \mathrm{s}\) . (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 \(\mathrm{m} ?\) (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ Assume that the tension of the string is constant and equal to \(W\) . (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for wayes traveling down the string?

Weightless Ant. Ant. An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu\) and is under tension \(F .\) Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wave-length \(\lambda\) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Work Energy and Power

Read Explanation

Electric Charge Field and Potential

Read Explanation

Medical Physics

Read Explanation

Astrophysics

Read Explanation

Translational Dynamics

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.