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Expert-verified Found in: Page 631 ### College Physics (Urone)

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 $$0.19{\rm{ }}m{\rm{ }}{s^{ - 1}}$$

See the step by step solution

## Step 1: Doppler effect in this case:

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

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

## Step 2:  Calculation for the velocity of blood

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 $$0.19{\rm{ }}m{\rm{ }}{s^{ - 1}}$$. ### Want to see more solutions like these? 