Laplace transform of $sin(sqrt 3t)$












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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?






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    0












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 20 at 5:07









      Dr.DoofusDr.Doofus

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          $begingroup$

          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.






          share|cite|improve this answer









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            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 20 at 5:12









                Dr.DoofusDr.Doofus

                12412




                12412






























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