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Chapter 12: Problem 51

Doppler ultrasound is used to measure the speed of blood flow (see Problem 42). The reflected sound interferes with the emitted sound, producing beats. If the speed of red blood cells is \(0.10 \mathrm{m} / \mathrm{s},\) the ultrasound frequency used is \(5.0 \mathrm{MHz},\) and the speed of sound in blood is \(1570 \mathrm{m} / \mathrm{s},\) what is the beat frequency?

Short Answer

Expert verified
Answer: The beat frequency is approximately \(1.598 \times 10^4 \; \mathrm{Hz}\).

Step by step solution

01

Write down the given information

The speed of red blood cells is \(v_r = 0.10 \; \mathrm{m} / \mathrm{s}\), the ultrasound frequency used is \(f_s = 5.0 \; \mathrm{MHz}\), and the speed of sound in blood is \(v_s = 1570 \; \mathrm{m} / \mathrm{s}\).
02

Convert the frequency to Hz

In order to use the units consistently, we need to convert the frequency from MHz to Hz: \(f_s = 5.0 \; \mathrm{MHz} = 5.0 \times 10^6 \; \mathrm{Hz}\).
03

Apply the Doppler effect formula for moving source

The Doppler effect formula for a moving source is \(f_r = \frac{v_s}{v_s \pm v_r} f_s\) where \(f_r\) is the frequency of the reflected wave, \(v_s\) is the speed of sound in the medium, \(v_r\) is the relative velocity of the source and \(\pm\) depends on whether the source is approaching or receding from the observer. In our case, the sound waves are being reflected off of the moving blood cells so we get: \(f_r = \frac{v_s}{v_s - v_r} f_s\).
04

Calculate the frequency of the reflected wave

Substitute the given values into the formula to find the frequency of the reflected wave: \(f_r = \frac{1570\; \mathrm{m} / \mathrm{s}}{1570\; \mathrm{m} / \mathrm{s} - 0.10 \; \mathrm{m} / \mathrm{s}} \times 5.0 \times 10^6 \; \mathrm{Hz} \approx 5.01598 \times 10^6 \; \mathrm{Hz}\).
05

Determine the beat frequency

The beat frequency is the difference between the frequencies of the emitted and reflected waves: \(f_b = |f_r - f_s| = |5.01598 \times 10^6 \; \mathrm{Hz} - 5.0 \times 10^6 \; \mathrm{Hz}| \approx 1.598 \times 10^4 \; \mathrm{Hz}\). The beat frequency due to the interference of the ultrasound reflected sound waves and the emitted sound waves is approximately \(1.598 \times 10^4 \; \mathrm{Hz}\).

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Most popular questions from this chapter

When playing fortissimo (very loudly), a trumpet emits sound energy at a rate of \(0.800 \mathrm{W}\) out of a bell (opening) of diameter \(12.7 \mathrm{cm} .\) (a) What is the sound intensity level right in front of the trumpet? (b) If the trumpet radiates sound waves uniformly in all directions, what is the sound intensity level at a distance of \(10.0 \mathrm{m} ?\)
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