Ttykuu

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

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

Can withdrawing asylum be illegal?

How can I protect witches in combat who wear limited clothing?

Finding degree of a finite field extension

Would it be possible to rearrange a dragon's flight muscle to somewhat circumvent the square-cube law?

Semisimplicity of the category of coherent sheaves?

Grover's algorithm - DES circuit as oracle?

Windows 10: How to Lock (not sleep) laptop on lid close?

I could not break this equation. Please help me

Sort list of array linked objects by keys and values

How many people can fit inside Mordenkainen's Magnificent Mansion?

Take groceries in checked luggage

Can smartphones with the same camera sensor have different image quality?

system call string length limit

Change bounding box of math glyphs in LuaTeX

First use of “packing” as in carrying a gun

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Match Roman Numerals

What's the point in a preamp?

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

Derivation tree not rendering

How to test the equality of two Pearson correlation coefficients computed from the same sample?

How are presidential pardons supposed to be used?

Why use ultrasound for medical imaging?

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

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

$endgroup$

share|cite|improve this question

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

share|cite|improve this question

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

edited 59 mins ago

Qmechanic♦

108k122001245

edited 59 mins ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

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.

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

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

edited 42 mins ago

answered 1 hour ago

tomtom

6,38711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  17 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
  

  
  
  
  

add a comment | 

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

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

Name

Email

Required, but never shown

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

edited 42 mins ago

answered 1 hour ago

tomtom

6,38711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  17 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
  

  
  
  
  

add a comment | 

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

edited 42 mins ago

answered 1 hour ago

tomtom

6,38711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  17 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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 42 mins ago

answered 1 hour ago

tomtom

6,38711627

$endgroup$

share|cite|improve this answer

edited 42 mins ago

answered 1 hour ago

tomtom

6,38711627

answered 1 hour ago

tomtom

6,38711627

answered 1 hour ago

tomtom

6,38711627

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844