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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 frequency required for a good scan is \(1.5×10^6 Hz\).

Step by step solution

01

Convert the wavelength to meters

The wavelength given in the exercise is in millimeters, but the speed is given in meters per second, so the wavelength needs to be converted to meters in order to use this in the wave equation. 1 mm is equal to 0.001 m, so the wavelength is \(0.001 m\).
02

Solve the wave equation for frequency

The wave equation is \(v = f \lambda\). Solve this equation for \(f\) to find: \(f = v / \lambda\).
03

Insert values and calculate frequency

Now that we have the equation for the frequency, we can insert the values for the speed and the wavelength: \(f = 1500 m/s / 0.001 m = 1.5×10^6 Hz\).

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Key Concepts

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

Sound Frequency
Sound frequency is a key concept in understanding how different types of sound are classified. Simply put, frequency is the number of times a sound wave's cycle occurs in one second, measured in hertz (Hz). The human ear can typically hear frequencies between 20 Hz and 20,000 Hz. Frequencies below 20 Hz are known as infrasound, while those above 20,000 Hz are referred to as ultrasound.

When we delve into the realm of ultrasound, we encounter frequencies that are too high-pitched for the human ear to detect. However, these frequencies are incredibly useful in medical diagnostics, particularly in ultrasound imaging. The physics behind ultrasound imaging rely on these high-frequency sound waves to penetrate the body and reflect off organs and tissues, creating images that can be used for medical analysis.
Ultrasound Frequency
Ultrasound frequencies, which typically range from 1 million to 15 million Hz (1 MHz to 15 MHz), are much higher than what our ears can perceive. In the context of medical imaging, the exact frequency chosen for an ultrasound scan affects the image quality. Higher frequencies provide better resolution because they produce shorter wavelengths, allowing for more detailed images of small structures.

In the case of our exercise, finding the right ultrasound frequency necessitates a trade-off between resolution and penetration depth. Higher frequencies do not travel as far into the body as lower frequencies, leading to lesser imaging depth. Therefore, medical professionals have to select an ultrasound frequency that is high enough to provide clear images and yet can penetrate sufficiently to reach the areas of interest. For the purposes of a detailed image with a wavelength of 1 mm, a frequency in the megahertz range, such as 1.5 MHz, is typically used.
Wave Equation
The wave equation is a fundamental principle in physics, connecting the speed (\(v\)), frequency (\(f\)), and wavelength (\( \text{\lambda} \text{(lambda)} \text{)} \)) of any wave. Mathematically, it is expressed as \(v = f \text{\lambda} \). For sound waves, this equation becomes critically important, as it allows us to calculate one of the variables if the other two are known.

Using the wave equation, ultrasound frequencies necessary for medical diagnostics can be determined. As done in the exercise's solution, you can rearrange the equation to solve for frequency when you are given the speed of sound in a medium (like tissue) and the desired wavelength for imaging. In the medical context, precise calculations are crucial to ensure that the ultrasound waves produce the best possible images while maintaining safety standards for patient exposure.

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Most popular questions from this chapter

Adjacent antinodes of a standing wave on a string are \(15.0 \mathrm{~cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude \(0.850 \mathrm{~cm}\) and period \(0.0750 \mathrm{~s}\). The string lies along the \(+x\) -axis and is fixed at \(x=0 .\) (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

A strong string of mass \(3.00 \mathrm{~g}\) and length \(2.20 \mathrm{~m}\) is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is \(9.00 \mathrm{~m} / \mathrm{s}\) The tension in the string is \(330 \mathrm{~N}\). (a) What is the amplitude of the standing wave at its antinode? (b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(M g\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu .\) You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M ;\) then you measure the node-to-node distance \(d\) for each standing-wave pattern and obtain the following data: $$ \begin{array}{l|lllll} \text { String } & \text { A } & \text { A } & \text { B } & \text { B } & \text { C } \\ \hline \mu(\mathrm{g} / \mathrm{cm}) & 0.0260 & 0.0260 & 0.0374 & 0.0374 & 0.0482 \\ M(\mathrm{~g}) & 559 & 249 & 365 & 207 & 262 \\ d(\mathrm{~cm}) & 48.1 & 31.9 & 32.0 & 24.2 & 23.8 \end{array} $$ (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^{2}(\) in \(\mathrm{kg} \cdot \mathrm{m})\) versus \(M(\) in \(\mathrm{kg}) .\) Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straightline fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). (d) For string A \((\mu=0.0260 \mathrm{~g} / \mathrm{cm}),\) what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of \(24.0 \mathrm{~cm}\) ? Use the value of \(f\) that you calculated in part (c).

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.

Tsunami! On December \(26,2004,\) a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed more than 200,000 people. Satellites observing these waves from space measured \(800 \mathrm{~km}\) from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) ? Does your answer help you understand why the waves caused such devastation?

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