Chapter 12, Problem 85GP

To determine

To Find: The beat frequency.

Answer to Problem 85GP

  $3\text{kHz}$

Explanation of Solution

Given:

The velocity of the blood flow, ${\mathrm{v}}_{\mathrm{b}}=0.32\text{m/s}$

The frequency of ultrasound wave, $\mathrm{f}=5.50\text{MHz}$

The speed of the wave, $\mathrm{v}=1.54\times {10}^3\text{m/s}$

Formula used:

By using the Doppler effect for the first wave

  $\mathrm{f}\text{'}=\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{b}}}{\mathrm{v}}\mathrm{f}$

Beat frequency for the second wave

  $\mathrm{f}\text{'}\text{'}=\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{b}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{b}}}\mathrm{f}\text{'}$

Here, ${\mathrm{v}}_{\mathrm{b}}$ is the velocity of the blood flow,

  $\mathrm{v}$ is the speed of the wave,

Write the expression for the beat frequency.

  $\mathrm{\Delta }\mathrm{f}=\mathrm{f}\text{'}-\mathrm{f}\text{'}\text{'}$

Calculation:

Write the expression of the frequency of the first wave.

  $\begin{array}{c}\mathrm{f}\text{'}=\frac{1540-0.32}{1540}\times 5.50\\ =5.498\text{MHz}\end{array}$

As a cell is the source of the sound of frequency $\mathrm{f}\text{'}$ and it moving away from the observer

using a formula of frequency

  $\begin{array}{c}\mathrm{f}\text{'}\text{'}=\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{s}}}\mathrm{f}\text{'}\\ =\frac{1540-0.32}{1540+0.32}\times 5.498\\ =5.495\text{MHz}\end{array}$

Write the expression for the beat frequency.

  $\begin{array}{c}\mathrm{\Delta }\mathrm{f}=\mathrm{f}\text{'}-\mathrm{f}\text{'}\text{'}\\ =5.498\text{MHz}-5.495\text{MHz}\\ =0.003\text{MHz}\\ =0.003\times {10}^6\text{Hz}\\ =3\times {10}^3\text{Hz}\\ =3\text{kHz}\end{array}$

Conclusion:

The beat frequency is $=3\text{kHz}$ .