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A \(2.00 \mathrm{MHz}\) 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 beats. The reflected sound is then mixed with the transmitted sound, and 85 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 \(31.875 \text{ m/s}\).

Step by step solution

01

Identify the Data Given

We know the following: - Frequency of the original sound wave (\(f_s\)) is \(2.00 \text{ MHz} = 2.00 \times 10^6 \text{ Hz}\).- The speed of sound in body tissue (\(v\)) is \(1500 \text{ m/s}\).- Beats detected per second (\(f_b\)) is \(85 \text{ Hz}\).
02

Understand the Doppler Effect

When a wave is reflected from a moving object, such as the fetal heart wall, a change in frequency will be observed due to the Doppler effect. The frequency detected by the listener (\(f'\)) relates to the source frequency (\(f_s\)) and the speed of the object (\(v_o\)) according to the Doppler shift formula.
03

Apply the Doppler Shift Formula for Moving Source

The formula for a reflective Doppler situation where the reflecting object moves toward the source at speed \(v_o\) is given by: \[ f' = \left(\frac{v + v_o}{v - v_o}\right) f_s\] where:- \(f'\) is the frequency of the reflected wave as perceived by the listener.- \(v\) is the speed of sound.- \(v_o\) is the speed of the fetal heart wall.
04

Calculate the Beat Frequency

The beat frequency \(f_b\) is the difference between the frequency of the reflected wave and the source frequency:\[ f_b = |f' - f_s|\]
05

Express the Relationship for Beat Frequency

Substitute the expression for \(f'\) from the Doppler effect formula into the beat frequency equation:\[ f_b = \left| \left(\frac{v + v_o}{v - v_o}\right) f_s - f_s \right|\]Since the heart wall is moving toward the sound receiver, we can assume a simple relationship \(f_b = (f' - f_s)\) because the reflected frequency is higher. This gives us:\[ f_b = f_s \left( \frac{v + v_o}{v - v_o} - 1 \right)\]
06

Solve for Velocity of the Heart Wall (\(v_o\))

Rearrange the equation to solve for the speed of the fetal heart wall \(v_o\):\[ 85 = 2.00 \times 10^6 \left( \frac{1500 + v_o}{1500 - v_o} - 1 \right)\]Simplify and solve this expression for \(v_o\). This requires solving a linear equation in terms of \(v_o\).
07

Simplify the Equation

Solve the equation by isolating the term containing \(v_o\) and simplifying:\[ 85 = 2.00 \times 10^6 \left( \frac{2v_o}{v - v_o} \right)\]Approximate and solve:\[ 85 \approx \frac{4.00 \times 10^6 v_o}{1500}\]
08

Find the Numerical Value

Solve for \(v_o\): \[ v_o \approx \frac{85 \times 1500}{4.00 \times 10^6}\]Calculate the numerical value to find the speed of the fetal heart wall.

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Key Concepts

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

Beat Frequency
When two sound waves of slightly different frequencies travel through the same medium and meet, they create a phenomenon known as a 'beat'. This is due to the interference of the two waves, leading to a fluctuation in sound intensity, which is perceived as a pulsing effect. The beat frequency is equal to the absolute difference between the frequencies of the two sound waves.
In the case of the fetal heart, when the sound wave reflects off the moving fetal heart wall, it results in two waves: the source frequency and the shifted frequency. The difference between these frequencies causes the beat frequency. Thus, the beat frequency heard, in this instance 85 Hz, indicates how much the frequency of the reflected wave differs from the original 2 MHz frequency.
It's a practical application of wave interference and can be crucial in measuring variations in frequency due to motion.
Sound Waves
Sound waves are a type of wave that travels through media such as air, water, or body tissue, carrying energy from a source to a receiver. These waves are characterized by their frequency, wavelength, and speed. The frequency is the number of wave cycles that pass a point in one second, measured in Hertz (Hz).
- **Frequency**: Determines the pitch of the sound.
- **Wavelength**: The distance between successive crests of the wave.
- **Amplitude**: Relates to the loudness of the sound.
In the context of the fetal heart measurement, a sound wave with a frequency of 2.00 MHz and a speed of 1500 m/s travels through the tissue. When this wave encounters the moving fetal heart wall, it gets reflected, and because the wall is moving, the reflected sound wave's frequency appears different, leading to the Doppler shift.
Reflective Doppler Shift
The Doppler effect occurs when a wave source moves relative to an observer, altering the frequency of waves reaching the observer compared to when both are stationary. This effect is observed with sound, light, and other types of waves.

When a sound wave reflects off a moving object, such as the fetal heart wall, we encounter a reflective Doppler shift. The change in frequency depends on the speed of the object and the speed of sound in the medium. In this situation, the heart wall moves towards the sound source, causing the frequency of the reflected wave to increase.
The Doppler shift formula used here: \[ f' = \left(\frac{v + v_o}{v - v_o}\right) f_s \]helps us calculate the observed frequency shift due to the motion of the heart wall. Here, \(f'\) is the frequency of the reflected wave as heard by the receiver, \(v\) is the speed of sound, and \(v_o\) is the speed at which the wall is moving. Understanding this shift is fundamental to estimating speeds based on changes in frequency.
Speed Calculation
Speed calculation in problems involving the Doppler effect can be tricky, but it's crucial for accurate measurement. After deriving the relationship between the frequencies of the source and the reflected sound wave, you can rearrange the formula to solve for the speed of the moving object, in this case, the fetal heart wall.

The equation: \[ 85 = 2.00 \times 10^6 \left( \frac{1500 + v_o}{1500 - v_o} - 1 \right) \]is simplified, allowing us to solve for \(v_o\), the speed of the fetal heart wall:- Rearrange terms to isolate \(v_o\), resulting in:\[ 85 \approx \frac{4.00 \times 10^6 v_o}{1500} \]By calculating and finding the numerical approximation, you determine the velocity \(v_o\), which is how fast the heart wall is moving towards the sound receiver. This process of solving provides insights into how physical equations apply to real-life medical diagnostics.

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Most popular questions from this chapter

Because the speed of ultrasound in bone is about twice the speed in tissue, the distance to a structure that lies beyond a bone may be measured incorrectly. If a beam passes through \(4 \mathrm{~cm}\) of tissue, then \(2 \mathrm{~cm}\) of bone, and then another \(1 \mathrm{~cm}\) of tissue before echoing off a cyst and returning to the transducer, what is the difference between the true distance to the cyst and the distance that's measured by assuming that the speed is always \(1540 \mathrm{~m} / \mathrm{s}\) ? Compared with the measured distance, the structure is actually A. \(1 \mathrm{~cm}\) farther. B. \(2 \mathrm{~cm}\) farther. C. I cm closer. D. \(2 \mathrm{~cm}\) closer.

A police siren of frequency \(f_{\text {siren }}\) is attached to a vibrating platform. The platform and siren oscillate up and down in simple harmonic motion with amplitude \(A_{\mathrm{p}}\) and frequency \(f_{\mathrm{p}}\). (a) Find the maximum and minimum sound frequencies that you would hear at a position directly above the siren. (b) At what point in the motion of the platform is the maximum frequency heard? The minimum frequency? Explain.

(a) Whale communication. Blue whales apparently communicate with each other using sound of frequency \(17 \mathrm{~Hz}\), which can be heard nearly \(1000 \mathrm{~km}\) away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is \(1531 \mathrm{~m} / \mathrm{s} ?\) (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength \(1.5 \mathrm{~cm}\) in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of \(25 \mathrm{kHz}\) for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide varicty of sounds, one type emits pulses of sound having a frequency between \(39 \mathrm{kHz}\) and \(78 \mathrm{kHz}\). What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as X-rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is \(1.0 \mathrm{~mm}\) across if the speed of sound in the tissue is \(1550 \mathrm{~m} / \mathrm{s} ?\)

(a) What is the sound intensity level in a car when the sound intensity is \(0.500 \mu \mathrm{W} / \mathrm{m}^{2} ?\) (b) What is the sound intensity in the air near a jackhammer when the sound intensity level is \(103 \mathrm{~dB}\) ?

While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of \(260 \mathrm{~Hz}\); your friend begins to blow his horn as well, and you hear a beat frequency of \(6.0 \mathrm{~Hz}\). How fast is your friend approaching you?

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