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Uses for doppler-shifted radar

Doppler-shifted radar echoes are used to measure wind velocities in storms as well as aircraft and automobile speeds. The principle is the same as for Doppler-shifted ultrasound. There is evidence that bats and dolphins may also sense the velocity of an object (such as prey) reflecting their ultrasound signals by observing its Doppler shift.

Calculate velocity of blood: doppler-shifted ultrasound

Ultrasound that has a frequency of 2.50 MHz is sent toward blood in an artery that is moving toward the source at 20.0 cm/s, as illustrated in [link] . Use the speed of sound in human tissue as 1540 m/s. (Assume that the frequency of 2.50 MHz is accurate to seven significant figures.)

  1. What frequency does the blood receive?
  2. What frequency returns to the source?
  3. What beat frequency is produced if the source and returning frequencies are mixed?
The picture represents an ultrasound device scanning the arteries and veins of a human hand.
Ultrasound is partly reflected by blood cells and plasma back toward the speaker-microphone. Because the cells are moving, two Doppler shifts are produced—one for blood as a moving observer, and the other for the reflected sound coming from a moving source. The magnitude of the shift is directly proportional to blood velocity.

Strategy

The first two questions can be answered using f obs = f s v w v w ± v s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {} and f obs = f s v w ± v obs v w size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right )} {} for the Doppler shift. The last question asks for beat frequency, which is the difference between the original and returning frequencies.

Solution for (a)

(1) Identify knowns:

  • The blood is a moving observer, and so the frequency it receives is given by
    f obs = f s v w ± v obs v w . size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right )} {}
  • v b size 12{v rSub { size 8{b} } } {} is the blood velocity ( v obs size 12{v rSub { size 8{"obs"} } } {} here) and the plus sign is chosen because the motion is toward the source.

(2) Enter the given values into the equation.

f obs = 2, 500 , 000 Hz 1540 m/s + 0 . 2 m/s 1540 m/s size 12{f rSub { size 8{"obs"} } = left (2,"500","000"" Hz" right ) left ( { {"1540"" m/s"+0 "." "2 m/s"} over {"1540 m/s"} } right )} {}

(3) Calculate to find the frequency: 2,500,325 Hz.

Solution for (b)

(1) Identify knowns:

  • The blood acts as a moving source.
  • The microphone acts as a stationary observer.
  • The frequency leaving the blood is 2,500,325 Hz, but it is shifted upward as given by
    f obs = f s v w v w v b . size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{b} } } } right )} {}

    f obs is the frequency received by the speaker-microphone.

  • The source velocity is v b size 12{v rSub { size 8{b} } } {} .
  • The minus sign is used because the motion is toward the observer.

The minus sign is used because the motion is toward the observer.

(2) Enter the given values into the equation:

f obs = 2, 500 , 325 Hz 1540 m/s 1540 m/s 0 . 200 m/s size 12{f rSub { size 8{"obs"} } = left (2,"500","325"" Hz" right ) left ( { {"1540"" m/s"} over {"1540 m/s " - 0 "." "200"" m/s"} } right )} {}

(3) Calculate to find the frequency returning to the source: 2,500,649 Hz.

Solution for (c)

(1) Identify knowns:

  • The beat frequency is simply the absolute value of the difference between f s size 12{f rSub { size 8{s} } } {} and f obs size 12{f rSub { size 8{"obs"} } } {} , as stated in:
    f B = f obs f s .

(2) Substitute known values:

2, 500 , 649 Hz 2, 500 , 000 Hz size 12{ lline 2,"500","649"`"Hz" - 2,"500","000"`"Hz" rline } {}

(3) Calculate to find the beat frequency: 649 Hz.

Discussion

The Doppler shifts are quite small compared with the original frequency of 2.50 MHz. It is far easier to measure the beat frequency than it is to measure the echo frequency with an accuracy great enough to see shifts of a few hundred hertz out of a couple of megahertz. Furthermore, variations in the source frequency do not greatly affect the beat frequency, because both f s size 12{f rSub { size 8{s} } } {} and f obs size 12{f rSub { size 8{"obs"} } } {} would increase or decrease. Those changes subtract out in f B = f obs f s .

Questions & Answers

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for the answer to complete, the units need specified why
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That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
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how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
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1.75cm
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Practice Key Terms 3

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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