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Ultrasonic waves of frequency \(4 \cdot 5 \mathrm{MHz}\) are used to detect tumour in soft tissues. The speed of sound in tissue is \(15 \mathrm{~km} \mathrm{~s}^{-1}\) and that in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\). Find the wavelength of this ultrasonic wave in air and in tissue.

Short Answer

Expert verified
Wavelength in air: \(7.56 \times 10^{-5} \, m\). Wavelength in tissue: \(3.33 \times 10^{-3} \, m\).

Step by step solution

01

Convert Frequencies

The frequency of the ultrasonic wave is given as \(4 \cdot 5 \, \text{MHz}\). First, convert this frequency to Hertz (Hz) for ease of calculation:\[ 4 \cdot 5 \, \text{MHz} = 4.5 \times 10^{6} \, \text{Hz} \]
02

Calculate Wavelength in Air

Use the formula for wavelength \[ \lambda = \frac{v}{f} \]where \(v\) is the speed of sound in air \(340 \, \text{m/s}\), and \(f\) is the frequency Substitute these values into the formula:\[ \lambda_{\text{air}} = \frac{340 \, \text{m/s}}{4.5 \times 10^{6} \, \text{Hz}} \approx 7.56 \times 10^{-5} \, \text{m} \]
03

Convert Speed of Sound in Tissue

The speed of sound in tissue is given as \(15 \, \text{km/s}\). Convert it to meters per second (m/s) to match the units with frequency:\[ 15 \, \text{km/s} = 15000 \, \text{m/s} \]
04

Calculate Wavelength in Tissue

Substitute the converted speed of sound in tissue into the wavelength formula:\[ \lambda_{\text{tissue}} = \frac{15000 \, \text{m/s}}{4.5 \times 10^{6} \, \text{Hz}} \approx 3.33 \times 10^{-3} \, \text{m} \]

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Key Concepts

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

Frequency
Frequency is a fundamental concept in understanding waves, including ultrasonic waves. It refers to the number of wave cycles that pass a given point in one second. The unit of frequency is Hertz (Hz), where one Hertz equals one cycle per second.
For ultrasonic waves, which are sound waves with frequencies above the human hearing range, this is particularly important. These waves have frequencies above 20 kHz. In our exercise, the given ultrasonic wave frequency is specified as 4.5 MHz, which is equivalent to 4.5 million cycles per second since 1 MHz equals 1 million Hz.
  • A high frequency like 4.5 MHz implies that the wave is oscillating very rapidly, completing 4,500,000 cycles in just one second.
  • Understanding the frequency allows us to understand the energy and penetration capability of the wave.
High-frequency waves are used in medical imaging, such as in the detection of tumors, because they can produce highly detailed images.
Speed of Sound
The speed of sound is how fast a sound wave travels through a medium. It can differ vastly depending on the medium—such as air, water, or tissue. For example, in air at room temperature, the speed of sound is roughly 340 meters per second (m/s).
In denser mediums like soft tissues, sound travels faster. In the given exercise, the speed of sound in tissue is 15 km/s, which translates to 15,000 m/s.
The difference in speed is due to the density and elasticity of the medium.
  • In general, sound travels more slowly in gases than in liquids and solid media.
  • Tissues in the body, being denser than air, transmit sound waves faster.
This variation in speed is crucial for applications in medical imaging, as it affects how and where waves are reflected or refracted, impacting the accuracy of the resulting imagery.
Wavelength Calculation
Wavelength is the distance between consecutive peaks of a wave. It's calculated using the formula: \[ \lambda = \frac{v}{f} \]where \( \lambda \) stands for wavelength, \( v \) is the speed of sound, and \( f \) is the frequency.
In the exercise, we are tasked with finding the wavelength of ultrasonic waves in both air and tissue.
- **In Air:** Given the speed of sound in air is 340 m/s and frequency 4.5 MHz, we can calculate the wavelength as: \[ \lambda_{\text{air}} = \frac{340 \text{ m/s}}{4.5 \times 10^{6} \text{ Hz}} \approx 7.56 \times 10^{-5} \text{ meters} \].

- **In Tissue:** With a speed of 15,000 m/s, the calculation goes: \[ \lambda_{\text{tissue}} = \frac{15000 \text{ m/s}}{4.5 \times 10^{6} \text{ Hz}} \approx 3.33 \times 10^{-3} \text{ meters} \].
  • Longer wavelengths can penetrate deeper but provide less detail in an image.
  • Shorter wavelengths enhance resolution, allowing more detailed imaging but may not penetrate as deeply.
Understanding how wavelength changes with different media is key not only in physics but also in applications like medical diagnostics, affecting the clarity and precision of ultrasound images.

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Most popular questions from this chapter

A boy riding on a bicycle going at \(12 \mathrm{~km} \mathrm{~h}^{-1}\) towards a vertical wall whistles at his dog on the ground. If the frequency of the whistle is \(1600 \mathrm{~Hz}\) and the speed of sound in air is \(330 \mathrm{~m} \mathrm{~s}^{-1}\), find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy.

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place \(6 \cdot 0 \mathrm{~m}\) from one of the speakers and \(6 \cdot 4 \mathrm{~m}\) from the other. If the sound signal is continuously varied from \(500 \mathrm{~Hz}\) to \(5000 \mathrm{~Hz}\), what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air \(=320 \mathrm{~m} \mathrm{~s}^{-1}\).

Three sources of sound \(S_{1}, S_{2}\) and \(S_{3}\) of equal intensity are placed in a straight line with \(S_{1} S_{2}=S_{2} S_{3}\) (figure 16-E5). At a point \(P\), far away from the sources, the wave coming from \(S_{2}\) is \(120^{\circ}\) ahead in phase of that from \(S_{1}\). Also, the wave coming from \(S_{3}\) is \(120^{\circ}\) ahead of that from \(S_{2}\). What would be the resultant intensity of sound at \(P ?\)

The separation between a node and the next antinode in a vibrating air column is \(25 \mathrm{~cm}\). If the speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\), find the frequency of vibration of the air column.

A small source of sound vibrating at frequency \(500 \mathrm{~Hz}\) is rotated in a circle of radius \(100 / \pi \mathrm{cm}\) at a constant angular speed of \(5 \cdot 0\) revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air \(=332 \mathrm{~m} \mathrm{~s}^{-1}\).

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