Why use ultrasound for medical imaging? The 2019 Stack Overflow Developer Survey Results Are...

Didn't get enough time to take a Coding Test - what to do now?

How to delete random line from file using Unix command?

Are my PIs rude or am I just being too sensitive?

Is a pteranodon too powerful as a beast companion for a beast master?

In horse breeding, what is the female equivalent of putting a horse out "to stud"?

How did passengers keep warm on sail ships?

ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?

Are spiders unable to hurt humans, especially very small spiders?

Why is the object placed in the middle of the sentence here?

Segmentation fault output is suppressed when piping stdin into a function. Why?

Sort list of array linked objects by keys and values

Do working physicists consider Newtonian mechanics to be "falsified"?

What are these Gizmos at Izaña Atmospheric Research Center in Spain?

Is every episode of "Where are my Pants?" identical?

What can I do if neighbor is blocking my solar panels intentionally?

What aspect of planet Earth must be changed to prevent the industrial revolution?

Am I ethically obligated to go into work on an off day if the reason is sudden?

University's motivation for having tenure-track positions

Why is superheterodyning better than direct conversion?

Simulating Exploding Dice

Road tyres vs "Street" tyres for charity ride on MTB Tandem

Relations between two reciprocal partial derivatives?

Make it rain characters

Change bounding box of math glyphs in LuaTeX



Why use ultrasound for medical imaging?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there an upper frequency limit to ultrasound?How is it possible for an Ultrasound device to correctly interpret a negative density change in tissue?How specifically does an MRI machine build an image from received radio wavesIs it possible to send modulated ultrasound wave from underwater to air?Why the Doppler Ultrasound beam needs to be looking directly down at a pipeIntensity of an ultrasound beam?Dynamic range of ultrasound machine expressed in dBUltrasound wave/beam generationDoppler effect of sound waves in bloodAveraging speed of ultrasound between two differnt boundaires












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?










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
    20 mins 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
    16 mins 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?










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
    20 mins 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
    16 mins 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?










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






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 59 mins ago









Qmechanic

108k122001245




108k122001245






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.









asked 1 hour ago









Ubaid HassanUbaid Hassan

19511




19511




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
    20 mins 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
    16 mins ago


















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
















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




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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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
    18 mins 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
    14 mins ago



















1












$begingroup$

Higher frequency provides higher resolution.






share|cite|improve this answer









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "151"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






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










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f472565%2fwhy-use-ultrasound-for-medical-imaging%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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
      18 mins 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
      14 mins ago
















    4












    $begingroup$

    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
      18 mins 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
      14 mins ago














    4












    4








    4





    $begingroup$

    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 42 mins ago

























    answered 1 hour ago









    tomtom

    6,38711627




    6,38711627












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











    1












    $begingroup$

    Higher frequency provides higher resolution.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Higher frequency provides higher resolution.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 1 hour 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


















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













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












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
















            Thanks for contributing an answer to Physics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f472565%2fwhy-use-ultrasound-for-medical-imaging%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Webac Holding Inhaltsverzeichnis Geschichte | Organisationsstruktur | Tochterfirmen |...

            What's the meaning of a knight fighting a snail in medieval book illustrations?What is the meaning of a glove...

            Salamanca Inhaltsverzeichnis Lage und Klima | Bevölkerungsentwicklung | Geschichte | Kultur und...