Equation of Simple Harmonic Oscillator

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So I know that the differential equation of a simple harmonic oscillator is



$dfracd^2xdt^2=-omega^2x$



and it's solution is $y = Asin(omega t+phi)$.



Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.



In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?



Edit:



Also when I start with the solution itself



$y=Asin(omega t + phi)$



$y'=Aomegacos(omega t + phi)$



$dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$



$omega^2cdot2yy'+2y'y''=0$



$omega^2=-dfracy''y$



$0=dfracyy'''-y''y'y^2$



$yy'''=y''y'$






calculus differential-equations







share|cite|improve this question



























    up vote
    2
    down vote

    favorite












    So I know that the differential equation of a simple harmonic oscillator is



    $dfracd^2xdt^2=-omega^2x$



    and it's solution is $y = Asin(omega t+phi)$.



    Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.



    In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?



    Edit:



    Also when I start with the solution itself



    $y=Asin(omega t + phi)$



    $y'=Aomegacos(omega t + phi)$



    $dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$



    $omega^2cdot2yy'+2y'y''=0$



    $omega^2=-dfracy''y$



    $0=dfracyy'''-y''y'y^2$



    $yy'''=y''y'$






    calculus differential-equations







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      So I know that the differential equation of a simple harmonic oscillator is



      $dfracd^2xdt^2=-omega^2x$



      and it's solution is $y = Asin(omega t+phi)$.



      Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.



      In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?



      Edit:



      Also when I start with the solution itself



      $y=Asin(omega t + phi)$



      $y'=Aomegacos(omega t + phi)$



      $dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$



      $omega^2cdot2yy'+2y'y''=0$



      $omega^2=-dfracy''y$



      $0=dfracyy'''-y''y'y^2$



      $yy'''=y''y'$






      calculus differential-equations







      share|cite|improve this question















      So I know that the differential equation of a simple harmonic oscillator is



      $dfracd^2xdt^2=-omega^2x$



      and it's solution is $y = Asin(omega t+phi)$.



      Now I learned for a solution with $n$ independent constants, the differential equation will be of order $n$.



      In this case, there are 3 independent constants, $A, omega, phi$, so shouldn't the differential equation be of order three? How is it two?



      Edit:



      Also when I start with the solution itself



      $y=Asin(omega t + phi)$



      $y'=Aomegacos(omega t + phi)$



      $dfracomega^2y^2A^2omega^2+dfracy'^2A^2w^2 = 1$



      $omega^2cdot2yy'+2y'y''=0$



      $omega^2=-dfracy''y$



      $0=dfracyy'''-y''y'y^2$



      $yy'''=y''y'$






      calculus differential-equations




      calculus differential-equations






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      share|cite|improve this question













      share|cite|improve this question




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      edited 2 hours ago

























      asked 2 hours ago









      Harshit Joshi

      319112




      319112




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.






          share|cite|improve this answer
















          • 1




            Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
            – Peter Szilas
            2 hours ago










          • @PeterSzilas Very nice Peter! Thanks a lot
            – gimusi
            2 hours ago










          • Gimusi.Your answer is nice!The link "links" omega and the spring constant.
            – Peter Szilas
            2 hours ago










          • But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
            – Harshit Joshi
            2 hours ago











          • @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
            – gimusi
            2 hours ago

















          up vote
          3
          down vote













          The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.






            share|cite|improve this answer
















            • 1




              Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
              – Peter Szilas
              2 hours ago










            • @PeterSzilas Very nice Peter! Thanks a lot
              – gimusi
              2 hours ago










            • Gimusi.Your answer is nice!The link "links" omega and the spring constant.
              – Peter Szilas
              2 hours ago










            • But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
              – Harshit Joshi
              2 hours ago











            • @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
              – gimusi
              2 hours ago














            up vote
            4
            down vote



            accepted










            Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.






            share|cite|improve this answer
















            • 1




              Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
              – Peter Szilas
              2 hours ago










            • @PeterSzilas Very nice Peter! Thanks a lot
              – gimusi
              2 hours ago










            • Gimusi.Your answer is nice!The link "links" omega and the spring constant.
              – Peter Szilas
              2 hours ago










            • But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
              – Harshit Joshi
              2 hours ago











            • @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
              – gimusi
              2 hours ago












            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.






            share|cite|improve this answer












            Note that $A$ and $phi$ are constants depending upon the initial conditions, which is consistent with the equation of order $2$, while $omega$ depends solely upon the system that is the given equation $ddot x=-omega x$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            gimusi

            77.6k73889




            77.6k73889







            • 1




              Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
              – Peter Szilas
              2 hours ago










            • @PeterSzilas Very nice Peter! Thanks a lot
              – gimusi
              2 hours ago










            • Gimusi.Your answer is nice!The link "links" omega and the spring constant.
              – Peter Szilas
              2 hours ago










            • But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
              – Harshit Joshi
              2 hours ago











            • @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
              – gimusi
              2 hours ago












            • 1




              Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
              – Peter Szilas
              2 hours ago










            • @PeterSzilas Very nice Peter! Thanks a lot
              – gimusi
              2 hours ago










            • Gimusi.Your answer is nice!The link "links" omega and the spring constant.
              – Peter Szilas
              2 hours ago










            • But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
              – Harshit Joshi
              2 hours ago











            • @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
              – gimusi
              2 hours ago







            1




            1




            Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
            – Peter Szilas
            2 hours ago




            Perhaps of interest,:hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
            – Peter Szilas
            2 hours ago












            @PeterSzilas Very nice Peter! Thanks a lot
            – gimusi
            2 hours ago




            @PeterSzilas Very nice Peter! Thanks a lot
            – gimusi
            2 hours ago












            Gimusi.Your answer is nice!The link "links" omega and the spring constant.
            – Peter Szilas
            2 hours ago




            Gimusi.Your answer is nice!The link "links" omega and the spring constant.
            – Peter Szilas
            2 hours ago












            But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
            – Harshit Joshi
            2 hours ago





            But like if I try to find the differential equation of the given solution by eliminating the constants, I land up at a third order equation $yy'''-y''y'$, so why is that?
            – Harshit Joshi
            2 hours ago













            @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
            – gimusi
            2 hours ago




            @HarshitJoshi differentiating and manipulating $ddot x=-omega x$ we obtain $dddot x=-omega dot x implies xdddot x=-omega xdot x =ddot xdot x implies xdddot x-ddot xdot x=0$ but that doesn't prove anything. The given ODE has 2 free constants which depends upon the initial condition $x(0)$ and $dot x(0)$.
            – gimusi
            2 hours ago










            up vote
            3
            down vote













            The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.






            share|cite|improve this answer
























              up vote
              3
              down vote













              The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.






                share|cite|improve this answer












                The constant $omega$ is given in the equation, you mustn't count it. What you count are the adjustable constants, that you can freely tune to achieve the initial conditions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Yves Daoust

                118k667214




                118k667214



























                     

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