Forced harmonic oscillator differential equation solution












0












$begingroup$


I have this differential equation :



$ddot{x} + omega_0^2x = F_0cos(omega t + Delta)$ with $omega_0^2 = k/m enspace .$



I take the complex equation



$ ddot{z} + omega_0^2 z = F_0e^{i(omega t + Delta)} enspace ,$



assuming an exponential solution : $alpha e^{ut + v}$



I find :



$z_p = alpha e^{ut + v} Leftrightarrow alpha u^2 e^{ut + v} + omega_0^2 alpha e^{ut + v} = F_0 e^{i(omega t + Delta)} $



$Leftrightarrow alpha e^{ut + v} = frac{F_0}{u^2 + omega_0^2} e^{i (omega t + Delta)}$



So $ z_p = frac{F_0}{omega_0^2 - omega^2}[cos(omega t + Delta) + i sin(omega t + Delta)]$



and I end up with :



$x_p = Re[z_p] = frac{F_0}{omega_0^2 - omega^2}cos(omega t + Delta) $



But the book states that a particular solution is :



$ x_p(t) = frac{F_0/m}{omega_0^2 - omega^2}cos(omega t + Delta) $



So where did the "m" come from ? Where did I go wrong ?






ordinary-differential-equations







share|cite|improve this question











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  • 1




    $begingroup$
    For me there is no $m$ in there, not even the units are consistent.
    $endgroup$
    – nicoguaro
    Sep 27 '14 at 16:50








  • 1




    $begingroup$
    Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
    $endgroup$
    – achille hui
    Sep 27 '14 at 17:05


















0












$begingroup$


I have this differential equation :



$ddot{x} + omega_0^2x = F_0cos(omega t + Delta)$ with $omega_0^2 = k/m enspace .$



I take the complex equation



$ ddot{z} + omega_0^2 z = F_0e^{i(omega t + Delta)} enspace ,$



assuming an exponential solution : $alpha e^{ut + v}$



I find :



$z_p = alpha e^{ut + v} Leftrightarrow alpha u^2 e^{ut + v} + omega_0^2 alpha e^{ut + v} = F_0 e^{i(omega t + Delta)} $



$Leftrightarrow alpha e^{ut + v} = frac{F_0}{u^2 + omega_0^2} e^{i (omega t + Delta)}$



So $ z_p = frac{F_0}{omega_0^2 - omega^2}[cos(omega t + Delta) + i sin(omega t + Delta)]$



and I end up with :



$x_p = Re[z_p] = frac{F_0}{omega_0^2 - omega^2}cos(omega t + Delta) $



But the book states that a particular solution is :



$ x_p(t) = frac{F_0/m}{omega_0^2 - omega^2}cos(omega t + Delta) $



So where did the "m" come from ? Where did I go wrong ?






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For me there is no $m$ in there, not even the units are consistent.
    $endgroup$
    – nicoguaro
    Sep 27 '14 at 16:50








  • 1




    $begingroup$
    Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
    $endgroup$
    – achille hui
    Sep 27 '14 at 17:05
















0












0








0





$begingroup$


I have this differential equation :



$ddot{x} + omega_0^2x = F_0cos(omega t + Delta)$ with $omega_0^2 = k/m enspace .$



I take the complex equation



$ ddot{z} + omega_0^2 z = F_0e^{i(omega t + Delta)} enspace ,$



assuming an exponential solution : $alpha e^{ut + v}$



I find :



$z_p = alpha e^{ut + v} Leftrightarrow alpha u^2 e^{ut + v} + omega_0^2 alpha e^{ut + v} = F_0 e^{i(omega t + Delta)} $



$Leftrightarrow alpha e^{ut + v} = frac{F_0}{u^2 + omega_0^2} e^{i (omega t + Delta)}$



So $ z_p = frac{F_0}{omega_0^2 - omega^2}[cos(omega t + Delta) + i sin(omega t + Delta)]$



and I end up with :



$x_p = Re[z_p] = frac{F_0}{omega_0^2 - omega^2}cos(omega t + Delta) $



But the book states that a particular solution is :



$ x_p(t) = frac{F_0/m}{omega_0^2 - omega^2}cos(omega t + Delta) $



So where did the "m" come from ? Where did I go wrong ?






ordinary-differential-equations







share|cite|improve this question











$endgroup$




I have this differential equation :



$ddot{x} + omega_0^2x = F_0cos(omega t + Delta)$ with $omega_0^2 = k/m enspace .$



I take the complex equation



$ ddot{z} + omega_0^2 z = F_0e^{i(omega t + Delta)} enspace ,$



assuming an exponential solution : $alpha e^{ut + v}$



I find :



$z_p = alpha e^{ut + v} Leftrightarrow alpha u^2 e^{ut + v} + omega_0^2 alpha e^{ut + v} = F_0 e^{i(omega t + Delta)} $



$Leftrightarrow alpha e^{ut + v} = frac{F_0}{u^2 + omega_0^2} e^{i (omega t + Delta)}$



So $ z_p = frac{F_0}{omega_0^2 - omega^2}[cos(omega t + Delta) + i sin(omega t + Delta)]$



and I end up with :



$x_p = Re[z_p] = frac{F_0}{omega_0^2 - omega^2}cos(omega t + Delta) $



But the book states that a particular solution is :



$ x_p(t) = frac{F_0/m}{omega_0^2 - omega^2}cos(omega t + Delta) $



So where did the "m" come from ? Where did I go wrong ?






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 27 '14 at 16:40









nicoguaro

1237




1237










asked Sep 27 '14 at 16:29









user1234161user1234161

290312




290312








  • 1




    $begingroup$
    For me there is no $m$ in there, not even the units are consistent.
    $endgroup$
    – nicoguaro
    Sep 27 '14 at 16:50








  • 1




    $begingroup$
    Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
    $endgroup$
    – achille hui
    Sep 27 '14 at 17:05
















  • 1




    $begingroup$
    For me there is no $m$ in there, not even the units are consistent.
    $endgroup$
    – nicoguaro
    Sep 27 '14 at 16:50








  • 1




    $begingroup$
    Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
    $endgroup$
    – achille hui
    Sep 27 '14 at 17:05










1




1




$begingroup$
For me there is no $m$ in there, not even the units are consistent.
$endgroup$
– nicoguaro
Sep 27 '14 at 16:50






$begingroup$
For me there is no $m$ in there, not even the units are consistent.
$endgroup$
– nicoguaro
Sep 27 '14 at 16:50






1




1




$begingroup$
Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
$endgroup$
– achille hui
Sep 27 '14 at 17:05






$begingroup$
Your starting equation should be $$ddot{x} + omega_0^2x = color{red}{frac{F_0}{m}}cos(omega t + Delta)$$ You have get rid of the mass $m$ in LHS, you need to do that to the RHS too.
$endgroup$
– achille hui
Sep 27 '14 at 17:05












1 Answer
1






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oldest

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$begingroup$

The general one degree-of-freedom motion equation for the mass-spring system has the form



$mddot{x}+kx=F(t)$



Dividing this equation by $m$, we get



$ddot{x}+omega_n^2x=F(t)/m=f(t)$



where $omega_n^2=frac{k}{m}$



For harmonic excitation, $F(t)=F_0cos(omega t+Delta)$. In standard form, then



$ddot{x}+omega_n^2x=f_0cos(omega t+Delta)$



The particular answer to this differential equation is



$x_p(t) = frac{f_0}{omega_n^2-omega^2}cos(omega t+Delta)$



So it all depends on how you write your force function. Either your equation should be



$ddot{x}+omega_n^2x=frac{F_0}{m}cos(omega t+Delta)$



to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.



This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $omega_n$ parameter) or general form (with the $m$ and $k$ parameters).






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    0












    $begingroup$

    The general one degree-of-freedom motion equation for the mass-spring system has the form



    $mddot{x}+kx=F(t)$



    Dividing this equation by $m$, we get



    $ddot{x}+omega_n^2x=F(t)/m=f(t)$



    where $omega_n^2=frac{k}{m}$



    For harmonic excitation, $F(t)=F_0cos(omega t+Delta)$. In standard form, then



    $ddot{x}+omega_n^2x=f_0cos(omega t+Delta)$



    The particular answer to this differential equation is



    $x_p(t) = frac{f_0}{omega_n^2-omega^2}cos(omega t+Delta)$



    So it all depends on how you write your force function. Either your equation should be



    $ddot{x}+omega_n^2x=frac{F_0}{m}cos(omega t+Delta)$



    to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.



    This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $omega_n$ parameter) or general form (with the $m$ and $k$ parameters).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The general one degree-of-freedom motion equation for the mass-spring system has the form



      $mddot{x}+kx=F(t)$



      Dividing this equation by $m$, we get



      $ddot{x}+omega_n^2x=F(t)/m=f(t)$



      where $omega_n^2=frac{k}{m}$



      For harmonic excitation, $F(t)=F_0cos(omega t+Delta)$. In standard form, then



      $ddot{x}+omega_n^2x=f_0cos(omega t+Delta)$



      The particular answer to this differential equation is



      $x_p(t) = frac{f_0}{omega_n^2-omega^2}cos(omega t+Delta)$



      So it all depends on how you write your force function. Either your equation should be



      $ddot{x}+omega_n^2x=frac{F_0}{m}cos(omega t+Delta)$



      to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.



      This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $omega_n$ parameter) or general form (with the $m$ and $k$ parameters).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The general one degree-of-freedom motion equation for the mass-spring system has the form



        $mddot{x}+kx=F(t)$



        Dividing this equation by $m$, we get



        $ddot{x}+omega_n^2x=F(t)/m=f(t)$



        where $omega_n^2=frac{k}{m}$



        For harmonic excitation, $F(t)=F_0cos(omega t+Delta)$. In standard form, then



        $ddot{x}+omega_n^2x=f_0cos(omega t+Delta)$



        The particular answer to this differential equation is



        $x_p(t) = frac{f_0}{omega_n^2-omega^2}cos(omega t+Delta)$



        So it all depends on how you write your force function. Either your equation should be



        $ddot{x}+omega_n^2x=frac{F_0}{m}cos(omega t+Delta)$



        to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.



        This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $omega_n$ parameter) or general form (with the $m$ and $k$ parameters).






        share|cite|improve this answer









        $endgroup$



        The general one degree-of-freedom motion equation for the mass-spring system has the form



        $mddot{x}+kx=F(t)$



        Dividing this equation by $m$, we get



        $ddot{x}+omega_n^2x=F(t)/m=f(t)$



        where $omega_n^2=frac{k}{m}$



        For harmonic excitation, $F(t)=F_0cos(omega t+Delta)$. In standard form, then



        $ddot{x}+omega_n^2x=f_0cos(omega t+Delta)$



        The particular answer to this differential equation is



        $x_p(t) = frac{f_0}{omega_n^2-omega^2}cos(omega t+Delta)$



        So it all depends on how you write your force function. Either your equation should be



        $ddot{x}+omega_n^2x=frac{F_0}{m}cos(omega t+Delta)$



        to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.



        This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $omega_n$ parameter) or general form (with the $m$ and $k$ parameters).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 19 '18 at 19:13









        ThalesThales

        214111




        214111






























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