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Q83PE

Expert-verifiedFound in: Page 631

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A diagnostic ultrasound echo is reflected from moving blood and returns with a frequency **\({\rm{500 Hz}}\)** higher than its original **\({\rm{2}}{\rm{.00 MHz}}\)**. What is the velocity of the blood? (Assume that the frequency of **\({\rm{2}}{\rm{.00 MHz}}\)** is accurate to seven significant figures and **\({\rm{500 Hz}}\)** is accurate to three significant figures.)**

The velocity of blood will be \(0.19{\rm{ }}m{\rm{ }}{s^{ - 1}}\)

**When there is a motion between observer and source then there will change in observed frequency due to that motion. **

Here in this case, there will be a double Doppler effect due to the motion of blood because that blood behaves like a source (when wave transmits) and observer (when wave reflects from the surface) and it is given as in mathematical form.

\(\begin{aligned} {f_{obs}} &= {f_s}\left( {\frac{{{v_w}}}{{{v_w} \pm {v_s}}}} \right)\left( {\frac{{{v_w} \pm {v_s}}}{{{v_w}}}} \right)\\ &= {f_s}\left( {\frac{{{v_w} \pm {v_s}}}{{{v_w} \pm {v_s}}}} \right)\end{aligned}\)

Here, \({v_s}\) is the speed of blood, \({v_w}\) is the speed of sound, \({f_{obs}}\) is observed frequency, and \({f_s}\) is the frequency of a source.

Now We know that blood will move towards the sound wave then we will get the higher observed frequency

\({f_{obs}} = {f_s}\left( {\frac{{{v_w} + {v_s}}}{{{v_w} - {v_s}}}} \right)\)

** **

Consider the given data as below.

The frequency \({f_{obs}}\) is \(500{\rm{ }}Hz\) higher than \({f_s}\). Therefore,

\(\begin{aligned} {f_{obs}} &= 2{\rm{ }}MHz + )\end{aligned}\)

Here, is the frequency of a source is \({f_s} = 2 \times {10^6}{\rm{ }}Hz\)The velocity of sound in blood, \({v_w} = 1570{\rm{ }}m{\rm{ }}{s^{ - 1}}\)

By putting these values into the formula we will get following

\(\begin{aligned} 500 + 2 \times {10^6}{\rm{ }} &= 2 \times {10^6}\left( {\frac{{1570 + {v_s}}}{{1570 - {v_s}}}} \right)\\\frac{{1.00025}}{1} &= \frac{{1570 + {v_s}}}{{1570 - {v_s}}}\\1.00025 \times (1570 - {v_s}) &= 1570 + {v_s}\end{aligned}\)

\(\begin{aligned} 1570.3985 - 1.00025{v_s} &= 1570 + {v_s}\\2.00025{v_s} &= 0.3985\end{aligned}\)

\(\begin{aligned} {v_s} &= \frac{{0.3985}}{{2.00025}}\\ &= 0.19{\rm{ }}m{\rm{ }}{s^{ - 1}}\end{aligned}\)

Hence, the velocity of blood will be \(0.19{\rm{ }}m{\rm{ }}{s^{ - 1}}\).

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