Solving differential equation describing motion in a pendulum












1












$begingroup$


I've been looking at Simple Harmonic Motion in particularly the period of a pendulum. This may seem like physics but my question is tailored towards mathematics. The differential equation is:



$${{d^2theta}over dt^2}+sintheta=0$$



Using the small angle approximation it is found that $T={2pi}sqrt{Lover g}$.



Is it possible to solve the differential equation without using the small angle approximation? If so, what is the actual period of a pendulum?






ordinary-differential-equations mathematical-physics







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
    $endgroup$
    – James Hanson
    Jan 31 '16 at 6:38










  • $begingroup$
    I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
    $endgroup$
    – Nikunj
    Jan 31 '16 at 6:45










  • $begingroup$
    Yeah essentially it will all boil down to elliptic integrals.
    $endgroup$
    – Triatticus
    Jan 31 '16 at 7:18
















1












$begingroup$


I've been looking at Simple Harmonic Motion in particularly the period of a pendulum. This may seem like physics but my question is tailored towards mathematics. The differential equation is:



$${{d^2theta}over dt^2}+sintheta=0$$



Using the small angle approximation it is found that $T={2pi}sqrt{Lover g}$.



Is it possible to solve the differential equation without using the small angle approximation? If so, what is the actual period of a pendulum?






ordinary-differential-equations mathematical-physics







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
    $endgroup$
    – James Hanson
    Jan 31 '16 at 6:38










  • $begingroup$
    I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
    $endgroup$
    – Nikunj
    Jan 31 '16 at 6:45










  • $begingroup$
    Yeah essentially it will all boil down to elliptic integrals.
    $endgroup$
    – Triatticus
    Jan 31 '16 at 7:18














1












1








1





$begingroup$


I've been looking at Simple Harmonic Motion in particularly the period of a pendulum. This may seem like physics but my question is tailored towards mathematics. The differential equation is:



$${{d^2theta}over dt^2}+sintheta=0$$



Using the small angle approximation it is found that $T={2pi}sqrt{Lover g}$.



Is it possible to solve the differential equation without using the small angle approximation? If so, what is the actual period of a pendulum?






ordinary-differential-equations mathematical-physics







share|cite|improve this question









$endgroup$




I've been looking at Simple Harmonic Motion in particularly the period of a pendulum. This may seem like physics but my question is tailored towards mathematics. The differential equation is:



$${{d^2theta}over dt^2}+sintheta=0$$



Using the small angle approximation it is found that $T={2pi}sqrt{Lover g}$.



Is it possible to solve the differential equation without using the small angle approximation? If so, what is the actual period of a pendulum?






ordinary-differential-equations mathematical-physics




ordinary-differential-equations mathematical-physics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 31 '16 at 6:35









JackJack

934




934












  • $begingroup$
    I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
    $endgroup$
    – James Hanson
    Jan 31 '16 at 6:38










  • $begingroup$
    I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
    $endgroup$
    – Nikunj
    Jan 31 '16 at 6:45










  • $begingroup$
    Yeah essentially it will all boil down to elliptic integrals.
    $endgroup$
    – Triatticus
    Jan 31 '16 at 7:18


















  • $begingroup$
    I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
    $endgroup$
    – James Hanson
    Jan 31 '16 at 6:38










  • $begingroup$
    I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
    $endgroup$
    – Nikunj
    Jan 31 '16 at 6:45










  • $begingroup$
    Yeah essentially it will all boil down to elliptic integrals.
    $endgroup$
    – Triatticus
    Jan 31 '16 at 7:18
















$begingroup$
I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
$endgroup$
– James Hanson
Jan 31 '16 at 6:38




$begingroup$
I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/…
$endgroup$
– James Hanson
Jan 31 '16 at 6:38












$begingroup$
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
$endgroup$
– Nikunj
Jan 31 '16 at 6:45




$begingroup$
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
$endgroup$
– Nikunj
Jan 31 '16 at 6:45












$begingroup$
Yeah essentially it will all boil down to elliptic integrals.
$endgroup$
– Triatticus
Jan 31 '16 at 7:18




$begingroup$
Yeah essentially it will all boil down to elliptic integrals.
$endgroup$
– Triatticus
Jan 31 '16 at 7:18










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let $ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.



begin{align*}
ddot{theta}+omega^{2} sin theta &= 0 \
omega &= sqrt{frac{mgell}{I}} \
k &= sqrt{frac{E}{2mg ell}}
end{align*}



$$
begin{array}{|c|c|c|c|} hline
& k < 1 & k = 1 & k > 1 \ hline
& & & \
displaystyle sin frac{theta}{2} &
koperatorname{sn} (omega t,k) &
tanh omega t &
displaystyle operatorname{sn} left( komega t,frac{1}{k} right) \
& & &\
theta &
2sin^{-1} (koperatorname{sn} (omega t,k)) &
4tan^{-1} e^{omega t}-pi &
displaystyle 2operatorname{am} left( komega t,frac{1}{k} right) \
& & &\
T &
displaystyle frac{4K(k)}{omega} &
infty &
displaystyle frac{2K(frac{1}{k})}{komega} \
& & &\ hline
end{array}$$



For small bob, $Iapprox mell^{2}$.



For $k<1$, amplitude $alpha=2sin^{-1} k$ and $k>1$ it's moving in complete circle.



For $k<<1$, $T=2pi sqrt{frac{I}{m gell}}
left(
1+frac{1}{4} sin^{2} frac{alpha}{2}+ldots
right)$



A plot of $T$ vs. $k$ with $omega=1$ is shown below



enter image description here






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    Let $ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.



    begin{align*}
    ddot{theta}+omega^{2} sin theta &= 0 \
    omega &= sqrt{frac{mgell}{I}} \
    k &= sqrt{frac{E}{2mg ell}}
    end{align*}



    $$
    begin{array}{|c|c|c|c|} hline
    & k < 1 & k = 1 & k > 1 \ hline
    & & & \
    displaystyle sin frac{theta}{2} &
    koperatorname{sn} (omega t,k) &
    tanh omega t &
    displaystyle operatorname{sn} left( komega t,frac{1}{k} right) \
    & & &\
    theta &
    2sin^{-1} (koperatorname{sn} (omega t,k)) &
    4tan^{-1} e^{omega t}-pi &
    displaystyle 2operatorname{am} left( komega t,frac{1}{k} right) \
    & & &\
    T &
    displaystyle frac{4K(k)}{omega} &
    infty &
    displaystyle frac{2K(frac{1}{k})}{komega} \
    & & &\ hline
    end{array}$$



    For small bob, $Iapprox mell^{2}$.



    For $k<1$, amplitude $alpha=2sin^{-1} k$ and $k>1$ it's moving in complete circle.



    For $k<<1$, $T=2pi sqrt{frac{I}{m gell}}
    left(
    1+frac{1}{4} sin^{2} frac{alpha}{2}+ldots
    right)$



    A plot of $T$ vs. $k$ with $omega=1$ is shown below



    enter image description here






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Let $ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.



      begin{align*}
      ddot{theta}+omega^{2} sin theta &= 0 \
      omega &= sqrt{frac{mgell}{I}} \
      k &= sqrt{frac{E}{2mg ell}}
      end{align*}



      $$
      begin{array}{|c|c|c|c|} hline
      & k < 1 & k = 1 & k > 1 \ hline
      & & & \
      displaystyle sin frac{theta}{2} &
      koperatorname{sn} (omega t,k) &
      tanh omega t &
      displaystyle operatorname{sn} left( komega t,frac{1}{k} right) \
      & & &\
      theta &
      2sin^{-1} (koperatorname{sn} (omega t,k)) &
      4tan^{-1} e^{omega t}-pi &
      displaystyle 2operatorname{am} left( komega t,frac{1}{k} right) \
      & & &\
      T &
      displaystyle frac{4K(k)}{omega} &
      infty &
      displaystyle frac{2K(frac{1}{k})}{komega} \
      & & &\ hline
      end{array}$$



      For small bob, $Iapprox mell^{2}$.



      For $k<1$, amplitude $alpha=2sin^{-1} k$ and $k>1$ it's moving in complete circle.



      For $k<<1$, $T=2pi sqrt{frac{I}{m gell}}
      left(
      1+frac{1}{4} sin^{2} frac{alpha}{2}+ldots
      right)$



      A plot of $T$ vs. $k$ with $omega=1$ is shown below



      enter image description here






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Let $ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.



        begin{align*}
        ddot{theta}+omega^{2} sin theta &= 0 \
        omega &= sqrt{frac{mgell}{I}} \
        k &= sqrt{frac{E}{2mg ell}}
        end{align*}



        $$
        begin{array}{|c|c|c|c|} hline
        & k < 1 & k = 1 & k > 1 \ hline
        & & & \
        displaystyle sin frac{theta}{2} &
        koperatorname{sn} (omega t,k) &
        tanh omega t &
        displaystyle operatorname{sn} left( komega t,frac{1}{k} right) \
        & & &\
        theta &
        2sin^{-1} (koperatorname{sn} (omega t,k)) &
        4tan^{-1} e^{omega t}-pi &
        displaystyle 2operatorname{am} left( komega t,frac{1}{k} right) \
        & & &\
        T &
        displaystyle frac{4K(k)}{omega} &
        infty &
        displaystyle frac{2K(frac{1}{k})}{komega} \
        & & &\ hline
        end{array}$$



        For small bob, $Iapprox mell^{2}$.



        For $k<1$, amplitude $alpha=2sin^{-1} k$ and $k>1$ it's moving in complete circle.



        For $k<<1$, $T=2pi sqrt{frac{I}{m gell}}
        left(
        1+frac{1}{4} sin^{2} frac{alpha}{2}+ldots
        right)$



        A plot of $T$ vs. $k$ with $omega=1$ is shown below



        enter image description here






        share|cite|improve this answer











        $endgroup$



        Let $ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.



        begin{align*}
        ddot{theta}+omega^{2} sin theta &= 0 \
        omega &= sqrt{frac{mgell}{I}} \
        k &= sqrt{frac{E}{2mg ell}}
        end{align*}



        $$
        begin{array}{|c|c|c|c|} hline
        & k < 1 & k = 1 & k > 1 \ hline
        & & & \
        displaystyle sin frac{theta}{2} &
        koperatorname{sn} (omega t,k) &
        tanh omega t &
        displaystyle operatorname{sn} left( komega t,frac{1}{k} right) \
        & & &\
        theta &
        2sin^{-1} (koperatorname{sn} (omega t,k)) &
        4tan^{-1} e^{omega t}-pi &
        displaystyle 2operatorname{am} left( komega t,frac{1}{k} right) \
        & & &\
        T &
        displaystyle frac{4K(k)}{omega} &
        infty &
        displaystyle frac{2K(frac{1}{k})}{komega} \
        & & &\ hline
        end{array}$$



        For small bob, $Iapprox mell^{2}$.



        For $k<1$, amplitude $alpha=2sin^{-1} k$ and $k>1$ it's moving in complete circle.



        For $k<<1$, $T=2pi sqrt{frac{I}{m gell}}
        left(
        1+frac{1}{4} sin^{2} frac{alpha}{2}+ldots
        right)$



        A plot of $T$ vs. $k$ with $omega=1$ is shown below



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 '18 at 12:55

























        answered Jan 31 '16 at 7:08









        Ng Chung TakNg Chung Tak

        14.8k31334




        14.8k31334






























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