Question

Ultrasound reflected from an oncoming bloodstream that is moving at30.0 cm/sis mixed with the original frequency of\({\bf{2}}.{\bf{50}}{\rm{ }}{\bf{MHz}}\)to produce beats. What is the beat frequency? (Assume that the frequency of\({\bf{2}}.{\bf{50}}{\rm{ }}{\bf{MHz}}\)is accurate to seven significant figures.)

Short answer

The beat frequency will be \(2003867{\rm{ }}Hz\).

Step-by-step solution

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 the bloodstream because that bloodstream 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, \({v_w}\)is the speed of blood, \({v_s}\)is the speed of sound, \({f_{obs}}\)is observed frequency/beat frequency, \({f_s}\)is the frequency of source/original frequency.

Now as known that the blood will move towards the sound wave then it will get a higher observed frequency like that.

\({f_{obs}} = {f_s}\left( {\frac{{{v_w} + {v_s}}}{{{v_w} - {v_s}}}} \right)\) ….. (1)

 Calculation for the velocity of blood:

Consider the given data as below.

The frequency of source/original frequency, \({f_s} = 2.5{\rm{ }}MHz = 2.5 \times {10^6}{\rm{ }}Hz\)

The velocity of sound in blood, \({v_w} = 1570{\rm{ }}m{\rm{ }}{s^{ - 1}}\)

The speed of source, \(v_{s}=30 cm/s=0.30 m/s\)

By putting these values into equation (1), you will get following.

\(\begin{aligned} {f_{obs}} &= 2 \times {10^6}\left( {\frac{{1570 + 0.03}}{{1570 - 0.03}}} \right)\\ &= 2 \times {10^6} \times 1.0019336\\ &= 2003867{\rm{ }}Hz\end{aligned}\)

Hence, the beat frequency will be \(2003867{\rm{ }}Hz\).