Why use ultrasound for medical imaging?












1












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What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?






energy acoustics frequency wavelength medical-physics







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  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    8 hours ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee♦
    8 hours ago










  • $begingroup$
    Homework-like question, lack of effort.
    $endgroup$
    – Pieter
    8 hours ago










  • $begingroup$
    you have a point, I’ll try to stop posting trivial questions from now on
    $endgroup$
    – Ubaid Hassan
    8 hours ago
















1












$begingroup$


What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?






energy acoustics frequency wavelength medical-physics







share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    8 hours ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee♦
    8 hours ago










  • $begingroup$
    Homework-like question, lack of effort.
    $endgroup$
    – Pieter
    8 hours ago










  • $begingroup$
    you have a point, I’ll try to stop posting trivial questions from now on
    $endgroup$
    – Ubaid Hassan
    8 hours ago














1












1








1





$begingroup$


What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?






energy acoustics frequency wavelength medical-physics







share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?






energy acoustics frequency wavelength medical-physics




energy acoustics frequency wavelength medical-physics






share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









Qmechanic♦

108k122001245




108k122001245






New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 9 hours ago









Ubaid HassanUbaid Hassan

30311




30311




New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    8 hours ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee♦
    8 hours ago










  • $begingroup$
    Homework-like question, lack of effort.
    $endgroup$
    – Pieter
    8 hours ago










  • $begingroup$
    you have a point, I’ll try to stop posting trivial questions from now on
    $endgroup$
    – Ubaid Hassan
    8 hours ago


















  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    8 hours ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee♦
    8 hours ago










  • $begingroup$
    Homework-like question, lack of effort.
    $endgroup$
    – Pieter
    8 hours ago










  • $begingroup$
    you have a point, I’ll try to stop posting trivial questions from now on
    $endgroup$
    – Ubaid Hassan
    8 hours ago
















$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
8 hours ago




$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
8 hours ago












$begingroup$
Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
$endgroup$
– dmckee♦
8 hours ago




$begingroup$
Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
$endgroup$
– dmckee♦
8 hours ago












$begingroup$
Homework-like question, lack of effort.
$endgroup$
– Pieter
8 hours ago




$begingroup$
Homework-like question, lack of effort.
$endgroup$
– Pieter
8 hours ago












$begingroup$
you have a point, I’ll try to stop posting trivial questions from now on
$endgroup$
– Ubaid Hassan
8 hours ago




$begingroup$
you have a point, I’ll try to stop posting trivial questions from now on
$endgroup$
– Ubaid Hassan
8 hours ago










3 Answers
3






active

oldest

votes


















8












$begingroup$

I think the simple answer here is resolution.



Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



$$lambda = {c over f} $$



so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



$$lambda = 0.001 {rm m} = 1 {rm mm}$$



At 20000 Hz $lambda = 75$ mm






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    8 hours ago






  • 1




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee♦
    8 hours ago



















3












$begingroup$

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).



The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.



And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.



Estimation



You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
$$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$



We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.




share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maybe "enter link description here" is not the link description you want.
    $endgroup$
    – dmckee♦
    8 hours ago



















2












$begingroup$

Higher frequency provides higher resolution.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      8 hours ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee♦
      8 hours ago
















    8












    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      8 hours ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee♦
      8 hours ago














    8












    8








    8





    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$



    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 9 hours ago

























    answered 9 hours ago









    tomtom

    6,42711627




    6,42711627












    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      8 hours ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee♦
      8 hours ago


















    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      8 hours ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee♦
      8 hours ago
















    $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    8 hours ago




    $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    8 hours ago




    1




    1




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee♦
    8 hours ago




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee♦
    8 hours ago











    3












    $begingroup$

    Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).



    The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



    On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
    $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
    where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.



    And as it is known,
    $$lambda = c/f,$$
    where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.



    Estimation



    You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
    $$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$



    We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.




    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Maybe "enter link description here" is not the link description you want.
      $endgroup$
      – dmckee♦
      8 hours ago
















    3












    $begingroup$

    Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).



    The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



    On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
    $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
    where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.



    And as it is known,
    $$lambda = c/f,$$
    where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.



    Estimation



    You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
    $$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$



    We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.




    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Maybe "enter link description here" is not the link description you want.
      $endgroup$
      – dmckee♦
      8 hours ago














    3












    3








    3





    $begingroup$

    Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).



    The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



    On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
    $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
    where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.



    And as it is known,
    $$lambda = c/f,$$
    where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.



    Estimation



    You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
    $$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$



    We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.




    share|cite|improve this answer











    $endgroup$



    Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).



    The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



    On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
    $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
    where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.



    And as it is known,
    $$lambda = c/f,$$
    where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.



    Estimation



    You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
    $$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$



    We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.





    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 7 hours ago

























    answered 8 hours ago









    DanielTuzesDanielTuzes

    21816




    21816












    • $begingroup$
      Maybe "enter link description here" is not the link description you want.
      $endgroup$
      – dmckee♦
      8 hours ago


















    • $begingroup$
      Maybe "enter link description here" is not the link description you want.
      $endgroup$
      – dmckee♦
      8 hours ago
















    $begingroup$
    Maybe "enter link description here" is not the link description you want.
    $endgroup$
    – dmckee♦
    8 hours ago




    $begingroup$
    Maybe "enter link description here" is not the link description you want.
    $endgroup$
    – dmckee♦
    8 hours ago











    2












    $begingroup$

    Higher frequency provides higher resolution.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Higher frequency provides higher resolution.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Higher frequency provides higher resolution.






        share|cite|improve this answer









        $endgroup$



        Higher frequency provides higher resolution.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        akhmeteliakhmeteli

        18.5k21844




        18.5k21844






















            Ubaid Hassan is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















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