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Speed of Heartwall in body

  1. Oct 9, 2005 #1
    A sound wave travels at a frequency 1.95 Mhz through a pregnant woman's abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 90.0 beats per second are detected. The speed of sound in body tissue is 1500 m/s.

    Here is what I came up with so far.

    (freq of beats) = [(speed of sound in body tissue)+(speed of listener, the reciever)]/[(speed of sound in body tissue)+(speed of source, the heart wall)] * (given freq sound wave) - (given freq sound wave)


    90 = (1500 +0)/(1500 + speed of heart wall) *1.95x10^6 - 1.95x10^6

    and I end up with a negative number -.06922

    What am Im doing wrong??

    Thank you for your help.

    I also tried just entering .06922 however the answer was still wrong.
  2. jcsd
  3. Oct 9, 2005 #2

    Andrew Mason

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    The beat frequency represents the difference between the frequencies of the incident and reflected sound waves. This is due to doppler shift. But you have to keep in mind that there are two doppler effects here. The heart wall is moving toward the source, which is equivalent to the source moving toward the heart wall, so the frequency which it 'hears' is doppler shifted up:

    [tex]f_1 = \frac{v}{v - v_{hw}}f_0[/tex]

    The reflected sound is again doppler shifted up because the source (the heart wall) is moving toward the observer:

    [tex]f_2 = \frac{v}{v - v_{hw}}f_1[/tex]

    It is the difference [itex]f_2-f_0 [/itex] that is detected as the beat frequency.

  4. Oct 9, 2005 #3
    I used the second equation that you have explained; however this time I get

    here is what I did

    1500/(1500+Vheart) * 1.95x10^6 = 90 + 1.95*10^6

    Am I using the equations incorrectly?
  5. Oct 9, 2005 #4

    Andrew Mason

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    No. You are using only one part of the doppler shift. There are two doppler shifts, effectively doubling the frequency shift.

    [tex]f_2 = \frac{v^2}{(v - v_{hw})^2}f_0[/tex]

    The speed of the heart wall should be about half of your figure.

    I would suggest that you develop the solution algebraically before plugging in numbers. It is easier to follow for you and for us.

  6. Oct 9, 2005 #5
    would f2 = the given 90 beats and f0 = the given frequency? 1.95x10^6? v = given speed of sound in body 1500 m/s

    after when i used those , i get an answer that is very very big.
  7. Oct 9, 2005 #6

    Andrew Mason

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    [tex]f_2 - f_0 = \frac{v^2}{(v - v_{hw})^2}f_0 - f_0[/tex]

    [tex]\Delta v = \sqrt{\frac{v^2}{(\Delta f/f_0 + 1)}}[/tex]

    [tex]\Delta v = \sqrt{\frac{1.5e3^2}{(90/1.95e6 + 1)}}[/tex]

    [tex]\Delta v = \sqrt{\frac{2.25e6}{(90/1950000 + 1)}[/tex]

    [tex]\Delta v = v - v_{hw} = 1499.965 m/sec[/tex]


    [tex]v_{hw} = 3.5e-2 = 3.5 cm/sec[/tex]

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