Mixing
Laplace transform of $sin(sqrt 3t)$
$begingroup$
This is an interesting question for me:
Find the Laplace transform of $mathcal{L} {sin(sqrt 3t)}$.
I know the Laplace transform of $sin(sqrt t) = frac{sqrtpi}{2s^{3/2}}e^{-1/4s}$, but simply multiplying t by 3 really messes things up.
I've looked through the books solution, and they write:
$mathcal{L} (sin(sqrt 3t)) = mathcal{L}(sin2(sqrtfrac{3}{4}t) = sqrt frac{3pi}{4} e^{-3/4s}{s^{3/2}}$. They say they do this by utilising the Laplace property of $mathcal{L}{frac{sin2sqrt {at}}{sqrt {pi a}}} = frac{e^{-a/s}}{s^{3/2}}$, which I don't understand (for example, there's no $sqrt pi a$ in the denominator, so what the heck?).
Can anyone please help explain how this works?
laplace-transform
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
This is an interesting question for me:
Find the Laplace transform of $mathcal{L} {sin(sqrt 3t)}$.
I know the Laplace transform of $sin(sqrt t) = frac{sqrtpi}{2s^{3/2}}e^{-1/4s}$, but simply multiplying t by 3 really messes things up.
I've looked through the books solution, and they write:
$mathcal{L} (sin(sqrt 3t)) = mathcal{L}(sin2(sqrtfrac{3}{4}t) = sqrt frac{3pi}{4} e^{-3/4s}{s^{3/2}}$. They say they do this by utilising the Laplace property of $mathcal{L}{frac{sin2sqrt {at}}{sqrt {pi a}}} = frac{e^{-a/s}}{s^{3/2}}$, which I don't understand (for example, there's no $sqrt pi a$ in the denominator, so what the heck?).
Can anyone please help explain how this works?
laplace-transform
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
This is an interesting question for me:
Find the Laplace transform of $mathcal{L} {sin(sqrt 3t)}$.
I know the Laplace transform of $sin(sqrt t) = frac{sqrtpi}{2s^{3/2}}e^{-1/4s}$, but simply multiplying t by 3 really messes things up.
I've looked through the books solution, and they write:
$mathcal{L} (sin(sqrt 3t)) = mathcal{L}(sin2(sqrtfrac{3}{4}t) = sqrt frac{3pi}{4} e^{-3/4s}{s^{3/2}}$. They say they do this by utilising the Laplace property of $mathcal{L}{frac{sin2sqrt {at}}{sqrt {pi a}}} = frac{e^{-a/s}}{s^{3/2}}$, which I don't understand (for example, there's no $sqrt pi a$ in the denominator, so what the heck?).
Can anyone please help explain how this works?
laplace-transform
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
$endgroup$
This is an interesting question for me:
Find the Laplace transform of $mathcal{L} {sin(sqrt 3t)}$.
I know the Laplace transform of $sin(sqrt t) = frac{sqrtpi}{2s^{3/2}}e^{-1/4s}$, but simply multiplying t by 3 really messes things up.
I've looked through the books solution, and they write:
$mathcal{L} (sin(sqrt 3t)) = mathcal{L}(sin2(sqrtfrac{3}{4}t) = sqrt frac{3pi}{4} e^{-3/4s}{s^{3/2}}$. They say they do this by utilising the Laplace property of $mathcal{L}{frac{sin2sqrt {at}}{sqrt {pi a}}} = frac{e^{-a/s}}{s^{3/2}}$, which I don't understand (for example, there's no $sqrt pi a$ in the denominator, so what the heck?).
Can anyone please help explain how this works?
laplace-transform
laplace-transform
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
asked Jan 20 at 5:07
Dr.DoofusDr.Doofus
12412
12412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I've figured it out.
It's just manipulating the expression, and then multiplying it by $sqrt frac{3pi}{4} $ to negate the denominator in the Laplace transform. Very smart.
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080208%2flaplace-transform-of-sin-sqrt-3t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I've figured it out.
It's just manipulating the expression, and then multiplying it by $sqrt frac{3pi}{4} $ to negate the denominator in the Laplace transform. Very smart.
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
I've figured it out.
It's just manipulating the expression, and then multiplying it by $sqrt frac{3pi}{4} $ to negate the denominator in the Laplace transform. Very smart.
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
I've figured it out.
It's just manipulating the expression, and then multiplying it by $sqrt frac{3pi}{4} $ to negate the denominator in the Laplace transform. Very smart.
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
$endgroup$
I've figured it out.
It's just manipulating the expression, and then multiplying it by $sqrt frac{3pi}{4} $ to negate the denominator in the Laplace transform. Very smart.
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
answered Jan 20 at 5:12
Dr.DoofusDr.Doofus
12412
12412
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080208%2flaplace-transform-of-sin-sqrt-3t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
