Question

The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) beat frequency of 240 Hz is observed. Assuming that the speed of sound in tissue is 1540 m/s, calculate the maximum velocity of the surface of the beating heart.

Answer 1

Answer:

$v = 2.88 \times 10^7 m/s$

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

$f' = f (\frac{v - v_h}{v + v_h})$

$f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})$

so we know that the beat frequency is

$\Delta f = 240 Hz$

so we will have

$f - f' = \Delta f$

$2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240$

$1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}$

so we have

$0.99989 = (\frac{v - 1540}{v + 1540})$

$1.99989\times 1540 = 1.067 \times 10^{-4} v$

$v = 2.88 \times 10^7 m/s$

Answer 2

The maximum velocity of the surface of the beating heart is 27,998,460 m/s.

Given the following data:

Maximum frequency = 240 Hz.

Frequency of ultrasound = $2.25 \times 10^{6}\;Hz$.

Speed of sound = 1540 m/s.

How to calculate maximum velocity of the surface.

Mathematically, Doppler's effect of waves is given by this formula:

$F_o = \frac{V-V_s}{V+V_s}F\\\\F_o = \frac{V-1540}{V+1540}\times 2.25 \times 10^{6}\;Hz$

For a frequency shift:

$Frequency \;shift = F-F_o \\\\240=(2.25 \times 10^{6})- \frac{V-1540}{V+1540}\times 2.25 \times 10^{6}$

Dividing all through by 2.25 \times 10^{6}, we have:

$1.07 \times 10^{-4}=1- \frac{V-1540}{V+1540}\\\\\frac{V-1540}{V+1540}=1-1.07 \times 10^{-4}\\\\\frac{V-1540}{V+1540}=0.99989\\\\V-1540=0.99989(V+1540)\\\\V-1540=0.99989V+1539.8306\\\\V-0.99989V=1540+1539.8306\\\\0.00011V=3079.8306\\\\V=\frac{3079.8306}{0.00011}$

V = 27,998,460 m/s.

Read more on Doppler's effect here: brainly.com/question/3841958