Solving differential equation describing motion in a pendulum
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?
differential-equations mathematical-physics
asked Jan 31 '16 at 6:35
Jack
934
add a comment |
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?
differential-equations mathematical-physics
asked Jan 31 '16 at 6:35
Jack
934
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/…
– James Hanson
Jan 31 '16 at 6:38
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18
add a comment |
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?
differential-equations mathematical-physics
asked Jan 31 '16 at 6:35
Jack
934
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?
differential-equations mathematical-physics
differential-equations mathematical-physics
asked Jan 31 '16 at 6:35
Jack
934
asked Jan 31 '16 at 6:35
Jack
934
asked Jan 31 '16 at 6:35
Jack
934
asked Jan 31 '16 at 6:35
Jack
934
asked Jan 31 '16 at 6:35
Jack
934
934
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/…
– James Hanson
Jan 31 '16 at 6:38
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18
add a comment |
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/…
– James Hanson
Jan 31 '16 at 6:38
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18
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/…
– James Hanson
Jan 31 '16 at 6:38
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/…
– James Hanson
Jan 31 '16 at 6:38
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18
add a comment |
1 Answer
1
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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
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
add a comment |
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
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
add a comment |
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
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
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
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
edited Jan 16 '18 at 12:55
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
answered Jan 31 '16 at 7:08
Ng Chung Tak
14.2k31334
14.2k31334
add a comment |
add a comment |
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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/…
– James Hanson
Jan 31 '16 at 6:38
I tried to find the expression for the trajectory, I believe it uses elliptical integrals.
– Nikunj
Jan 31 '16 at 6:45
Yeah essentially it will all boil down to elliptic integrals.
– Triatticus
Jan 31 '16 at 7:18