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

What was the last x86 CPU that did not have the x87 floating-point unit built in?

How to delete random line from file using Unix command?

Typeface like Times New Roman but with "tied" percent sign

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

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

Python - Fishing Simulator

Take groceries in checked luggage

Who or what is the being for whom Being is a question for Heidegger?

How do you keep chess fun when your opponent constantly beats you?

Was credit for the black hole image misattributed?

Can a novice safely splice in wire to lengthen 5V charging cable?

Derivation tree not rendering

Why not take a picture of a closer black hole?

Does the AirPods case need to be around while listening via an iOS Device?

Change bounding box of math glyphs in LuaTeX

Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?

Scientific Reports - Significant Figures

How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?

The variadic template constructor of my class cannot modify my class members, why is that so?

Finding degree of a finite field extension

He got a vote 80% that of Emmanuel Macron’s

How to prevent selfdestruct from another contract

"... to apply for a visa" or "... and applied for a visa"?

How do I add random spotting to the same face in cycles?



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
    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
    12 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
    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
    12 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 55 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
    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
    12 mins ago


















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
















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




$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
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
12 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
12 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
    13 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
    9 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
      13 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
      9 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
      13 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
      9 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 38 mins ago

























    answered 57 mins 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
      13 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
      9 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
      13 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
      9 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
    13 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
    13 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
    9 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
    9 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 57 mins 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

            “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

            How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

            變成蝙蝠會怎樣? 參考資料 外部連結 导航菜单Thomas Nagel, "What is it like to be a...