# Solving for the eqs of motion for a Double Pendulum using a Lagrangian

• Wavefunction
In summary, the conversation discusses a problem involving two masses connected by a rigid rod and two strings of different lengths attached to a fixed point. The equation of motion for the system is derived using Lagrangian mechanics and the frequency of small oscillations around the equilibrium position is found. The difference between an "extensionless string" and a "string" is also discussed, with the former emphasizing the constraint of the particle's movement on a fixed radius from the fixed point. The possibility of simplifying the problem by setting the lengths and masses to be equal is also mentioned, but the focus of the homework is to practice using Lagrangian mechanics.
Wavefunction

## Homework Statement

Two masses $m_1$ and $m_2 (m_1 ≠ m_2)$ are connected by a rigid rod of length $d$ and of negligible
mass. An extensionless string of length$l_1$ is attached to $m_1$ and connected to a fixed point $P$.
Similarly, a string of length $l_2 (l_1 ≠ l_2)$ connects $m_2$ and $P$. Obtain the equation describing the
motion in the plane of $m_1$, $m_2$, and $P$ and find the frequency of small oscillations around the
equilibrium position.

## Homework Equations

(1) $\frac{∂L}{∂q_j}-\frac{d}{dt}\frac{∂L}{∂\dot{q_j}}+\sum_kλ_k(t)\frac{∂f_k(q_j,t)}{∂q_j} = 0$

(2) $s=Dn-m$

## The Attempt at a Solution

Okay so the first thing I did in this problem was I tried to figure out how many coordinates I need. Each particle can move in 2 dimensions so $D=2$. There are 2 particles so $n=2$. There are 3 constraints:(a) $x_1^2+y_1^2 = l_1^2$, (b) $x_2^2+y_2^2 = l_1^2$, and (c) $(x_2-x_1)^2+(y_2-y_1)^2 = d^2$

$s=(2*2)-3 = 1$ However, I'll start of with the original four coordinates and work my way to only 1:

The Lagrangian:
$L=T-U → T=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]$ and $U=g[m_1y_1+m_2y_2]$
Then $L=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]-g[m_1y_1+m_2y_2]$

Now I can apply 2 out of my 3 constraints (a and b):
$x_1 = l_1sin(θ_1)$
$y_1 = l_1cos(θ_1)$
$x_2 = l_2sin(θ_2)$
$y_2 = l_2cos(θ_2)$

Applying these transformations to the Lagrangian:

$L=\frac{1}{2}[m_1l_1^2\dot{θ_1}^2+m_2l_2^2\dot{θ_2}^2]-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)]$

Now the angular velocity $\dot{θ}$ is the same for both particles because the thin massless rod of length $d$ holds them together. So $\dot{θ_1}=\dot{θ_2}$

Applying this to L:

$L=\frac{1}{2}[m_1l_1^2+m_2l_2^2]\dot{θ_1}^2-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)]$

$\dot{θ_1}=\dot{θ_2} → θ_1=θ_2+C → f = θ_1-θ_2-C=0$

Okay now I can use eq. (1):

$θ_1$: $\frac{∂L}{∂θ_1} = g[m_1l_1sin(θ_1)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_1}}] = \frac{d}{dt}[[m_1l_1^2+m_2l_2^2]\dot{θ_1}]$, and $λ\frac{∂f}{∂θ_1} = λ$

(3) $g[m_1l_1sin(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}+λ=0$

$θ_2$: $\frac{∂L}{∂θ_2} = g[m_2l_2sin(θ_2)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_2}}] = \frac{d}{dt}[0]$, and $λ\frac{∂f}{∂θ_1} = -λ$

(4) $g[m_2l_2sin(θ_2)]-λ=0$

From eq. 4 I know $λ$ in terms of $θ_1$:

Using $θ_1-C=θ_2$; $λ=g[m_2l_2sin(θ_1-C)]$

Then eq. 3 becomes:

$g[m_1l_1sin(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}+g[m_2l_2sin(θ_1-C)]=0$

$g[m_1l_1sin(θ_1)]+g[m_2l_2[sin(θ_1)cos(C)-sin(C)cos(θ_1)]]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0$

$[gm_1l_1+gm_2l_2cos(C)]sin(θ_1)-gm_2l_2sin(C)cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0$

Now there is the matter of $C$ to handle:

If I plug in the coordinate transforms into constraint (c) I will arrive at the law of cosines:

$l_1^2+l_2^2-2l_1l_2cos(θ_1-θ_2) = d^2$ (keep in mind here that I have defined $θ_1 < 0$ and $θ_2>0$)

From the law of cosines I get:

$θ_1 = θ_2+arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})$ so I have now found what C is:

$C = arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})$

Now plugging in $C$ into the new form of eq. 3:

$[gm_1l_1+gm_2l_2cos(arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}))]sin(θ_1)-gm_2l_2sin(arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}))cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0$

$[gm_1l_1+gm_2l_2(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})]sin(θ_1)-gm_2l_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{-2l_1l_2})cos(θ_1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0$

Okay so now I only want to consider small oscillations about the equilibrium position $\Rightarrow θ_1\ll 1$

Then eq. 3 becomes the Linear ODE:

$g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})](θ_1)+gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})(1)-[m_1l_1^2+m_2l_2^2]\ddot{θ_1}=0$

Now this equation can be put into the form $\ddot{θ}+ω_0^2θ=f(t)$ where $f(t)$ is a constant driving force.

$\ddot{θ_1}+\frac{-g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})]}{m_1l_1^2+m_2l_2^2}θ_1=\frac{gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})}{m_1l_1^2+m_2l_2^2}$

Now then $ω_0^2 \equiv \frac{-g[m_1l_1-m_2(\frac{d^2-l_1^2-l_2^2}{2l_1})]}{m_1l_1^2+m_2l_2^2}$ (frequency of oscillation about the equilibrium point)

The equilibrium condition is given as $θ_1(0) = \frac{f(t)}{ω_0^2}$ because in equilibrium $\ddot{θ_1(t=0)}=0$ In this case $f(t) = \frac{gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1})}{m_1l_1^2+m_2l_2^2}$

The general solution to eq 3. is:

(4) $θ_1(t) = Acos(ω_0t)+Bsin(ω_0t)+θ_p(t)$

$\ddot{θ_p}+ω_0^2θ_p = gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2l_1}) → θ_p = gm_2(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{2ω_0^2l_1})$

$A = θ_1(0)-θ_p$ then eq. 4 becomes $θ_1(t) = [θ_1(0)-θ_p]cos(ω_0)+Bsin(ω_0t)+θ_p$

$\dot{θ_1} = Bω_0cos(ω_0t) → \dot{θ_1(0)} = Bω_0 → B=\frac{ \dot{θ_1(0)}}{ω_0}$

So the general solution is then:

$θ_1(t) = [θ_1(0)-θ_p]cos(ω_0t)+\frac{ \dot{θ_1(0)}}{ω_0}sin(ω_0t)+θ_p$

Thank you in advance for helping me out in checking my work(:

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What is the difference between and "extensionless string" and a "string" ?

maajdl said:
What is the difference between and "extensionless string" and a "string" ?

In the context of the problem it is simply emphasizing the point that the particle is constrained to move on a fixed radius $l_i$ from point
$P$

Does that mean that there is no difference between a rod and a string? (in this problem)

If that is true, then the triangle (l1 ,l2 ,d) is a rigid triangle, and then the problem admits an almost trivial solution: a simple pendulum whose you need to find the parameters.
If that is true, you might be able to check your solution directly from this point of view.

However, I still have some doubts about the meaning of the question.

maajdl said:
Does that mean that there is no difference between a rod and a string? (in this problem)

If that is true, then the triangle (l1 ,l2 ,d) is a rigid triangle, and then the problem admits an almost trivial solution: a simple pendulum whose you need to find the parameters.
If that is true, you might be able to check your solution directly from this point of view.

However, I still have some doubts about the meaning of the question.

Yes, I suppose I could set $l_1= l_2$ and $m_1 = m_2$ because in that limit I should be able to calculate the reduced mass $\mu$ and check that the equations of motion I came up with match that of the simple pendulum. However, the point of this homework is to exercise one's skill with using Lagrangian mechanics. I really want to make sure my setup in this regard is correct more so than the equations of motion per se. Thank you for the suggestion though(:

I haven't gotten through all of your manipulations. But I did notice a couple of things.

1) In getting to equations (3) and (4), you should still be treating ##\theta_1## and ##\theta_2## as independent variables in the Lagrange multiplier method.

So, ##\frac{\partial L}{\partial \theta_2} \neq 0##. [EDIT: As discussed in the following posts, this should have been ##\frac{\partial L}{\partial \dot \theta_2} \neq 0##]

2) The small oscillation approximation does not mean that ##\theta_1## (or ##\theta_2##) is small. When the system hangs at rest in equilibrium, ##\theta_1## has some value ##\theta_1^0## which is not necessarily small. It's the deviation of ##\theta_1## from ##\theta_1^0## that will be small.

--------------------
Note that as a check, you can obtain the frequency of oscillation by considering the system as a physical pendulum and using the well-known formula for the frequency of a physical pendulum.

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1 person
TSny said:
I haven't gotten through all of your manipulations. But I did notice a couple of things.

1) In getting to equations (3) and (4), you should still be treating ##\theta_1## and ##\theta_2## as independent variables in the Lagrange multiplier method.

So, ##\frac{\partial L}{\partial \theta_2} \neq 0##.

2) The small oscillation approximation does not mean that ##\theta_1## (or ##\theta_2##) is small. When the system hangs at rest in equilibrium, ##\theta_1## has some value ##\theta_1^0## which is not necessarily small. It's the deviation of ##\theta_1## from ##\theta_1^0## that will be small.

--------------------
Note that as a check, you can obtain the frequency of oscillation by considering the system as a physical pendulum and using the well-known formula for the frequency of a physical pendulum.

In 1) were you referring to $\frac{∂L}{∂\dot{θ_2}}$ because I never got that $\frac{∂L}{∂θ_2}= 0$ I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations. What about $\dot{θ_1}=\dot{θ_2}$? Is that correct? I got that relation 2 different ways. The first was from the law of cosines relation and the second was treating the two particles as having a constant separation $d$ in the plane.

In 2) So basically I need to reformulate the solution to 3)?

Wavefunction said:
In 1) were you referring to $\frac{∂L}{∂\dot{θ_2}}$ because I never got that $\frac{∂L}{∂θ_2}= 0$ I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations. What about $\dot{θ_1}=\dot{θ_2}$? Is that correct? I got that relation 2 different ways. The first was from the law of cosines relation and the second was treating the two particles as having a constant separation $d$ in the plane.

In 2) So basically I need to reformulate the solution to 3)?

Ahhh yes I see, wow I can't believe I overlooked that $ε$ the angular displacement from $θ_1(0)$ is what is small. Thank you for pointing that out kind sir!

Wavefunction said:
In 1) were you referring to $\frac{∂L}{∂\dot{θ_2}}$ because I never got that $\frac{∂L}{∂θ_2}= 0$

Yes, sorry. I should have written ##\frac{∂L}{∂\dotθ_2}\neq 0##

I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations.

As I understand it, when using the Lagrange multiplier method you shouldn't use your constraint conditions until after you've derived the equations of motion from the Lagrangian (treating ##\theta_1## and ##\theta_2## as independent variables).

What about $\dot{θ_1}=\dot{θ_2}$? Is that correct?

Yes. But you shouldn't use it until after you have derived the correct equations of motion.

1 person
Wavefunction said:

## Homework Statement

Two masses $m_1$ and $m_2 (m_1 ≠ m_2)$ are connected by a rigid rod of length $d$ and of negligible
mass. An extensionless string of length$l_1$ is attached to $m_1$ and connected to a fixed point $P$.
Similarly, a string of length $l_2 (l_1 ≠ l_2)$ connects $m_2$ and $P$. Obtain the equation describing the
motion in the plane of $m_1$, $m_2$, and $P$ and find the frequency of small oscillations around the
equilibrium position.

## Homework Equations

(1) $\frac{∂L}{∂q_j}-\frac{d}{dt}\frac{∂L}{∂\dot{q_j}}+\sum_kλ_k(t)\frac{∂f_k(q_j,t)}{∂q_j} = 0$

(2) $s=Dn-m$

## The Attempt at a Solution

Okay so the first thing I did in this problem was I tried to figure out how many coordinates I need. Each particle can move in 2 dimensions so $D=2$. There are 2 particles so $n=2$. There are 3 constraints:(a) $x_1^2+y_1^2 = l_1^2$, (b) $x_2^2+y_2^2 = l_1^2$, and (c) $(x_2-x_1)^2+(y_2-y_1)^2 = d^2$

$s=(2*2)-3 = 1$ However, I'll start of with the original four coordinates and work my way to only 1:

The Lagrangian:
$L=T-U → T=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]$ and $U=g[m_1y_1+m_2y_2]$
Then $L=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]-g[m_1y_1+m_2y_2]$

Now I can apply 2 out of my 3 constraints (a and b):
$x_1 = l_1sin(θ_1)$
$y_1 = l_1cos(θ_1)$
$x_2 = l_2sin(θ_2)$
$y_2 = l_2cos(θ_2)$

Applying these transformations to the Lagrangian:

$L=\frac{1}{2}[m_1l_1^2\dot{θ_1}^2+m_2l_2^2\dot{θ_2}^2]-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)]$

Now applying the transformations to constraint (c) $l_1^2+l_2^2 -2l_1l_2cos(θ_1-θ_2)=d^2$ This will be my constraint in (1)

Okay now I can use eq. (1):

$θ_1$: $\frac{∂L}{∂θ_1} = g[m_1l_1sin(θ_1)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_1}}] = \frac{d}{dt}[[m_1l_1^2]\dot{θ_1}]$, and $λ\frac{∂f}{∂θ_1} = λ(2l_1l_2sin(θ_1-θ_2))$

(3) $g[m_1l_1sin(θ_1)]-[m_1l_1^2]\ddot{θ_1}+λ(2l_1l_2sin(θ_1-θ_2))=0$

$θ_2$: $\frac{∂L}{∂θ_2} = g[m_2l_2sin(θ_2)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_2}}] = \frac{d}{dt}[[m_2l_2^2]\dot{θ_2}]$, and $λ\frac{∂f}{∂θ_1} = -λ(2l_1l_2sin(θ_1-θ_2))$

(4) $g[m_2l_2sin(θ_2)]-[m_2l_2^2]\ddot{θ_2}-λ(2l_1l_2sin(θ_1-θ_2))=0$

Okay now I can use the fact that $\dot{θ}$ is the same for both particles:

$\dot{θ_1}=\dot{θ_2}$, $\ddot{θ_1}=\ddot{θ_2}$, and $θ_1= θ_2+C$

Then (3) and (4) become:

(5)$g[m_1l_1sin(θ_1)]-[m_1l_1^2]\ddot{θ_1}+λ(2l_1l_2sin(C))=0$

(6)$g[m_2l_2sin(θ_1-C)]-[m_2l_2^2]\ddot{θ_1}-λ(2l_1l_2sin(C))=0$

Now adding eqs (5) and (6) together:

$g[m_1l_1sin(θ_1)+m_2l_2sin(θ_1-C)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

$g[m_1l_1sin(θ_1)+m_2l_2[sin(θ_1)cos(C)-sin(C)cos(θ_1)]]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

(7) $g[[m_1l_1+m_2l_2cos(C)]sin(θ_1)-m_2l_2sin(C)cos(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

Now if the pendulum is in static equilibrium then $θ_1^0$ is a constant which implies $\ddot{θ_1^0} = 0$. Now if I consider some small angular perturbation $ε(t)$ about $θ_1$ Also $θ_1(t)= θ_1^0+ε(t)$ then eq (7) becomes:

(8) $g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0+ε)-m_2l_2sin(C)cos(θ_1^0+ε)]-[m_1l_1^2+m_2l_2^2][\ddot{θ_1^0}+\ddot{ε}] = 0$

$g[[m_1l_1+m_2l_2cos(C)][sin(ε)cos(θ_1^0)+sin(θ_1^0)cos(ε)]-m_2l_2sin(C)[cos(θ_1^0)cos(ε)-sin(θ_1^0)sin(ε)]]-[m_1l_1^2+m_2l_2^2]\ddot{ε} = 0$

Now since the oscillations are small $sin(ε)≈ε$ and $cos(ε)≈1$

$g[[m_1l_1+m_2l_2cos(C)][εcos(θ_1^0)+sin(θ_1^0)]-m_2l_2sin(C)[cos(θ_1^0)-sin(θ_1^0)ε]]-[m_1l_1^2+m_2l_2^2]\ddot{ε} = 0$

$g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0)-m_2l_2sin(C)cos(θ_1^0) = [m_1l_1^2+m_2l_2^2]\ddot{ε}+[-g[[m_1l_1+m_2l_2cos(C)]cos(θ_1^0)-m_2l_2sin(C)sin(θ_1^0)]ε$

$\ddot{ε}+ω^2ε=α(t)$ where $α(t) = \frac{g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0)-m_2l_2sin(C)cos(θ_1^0)}{m_1l_1^2+m_2l_2^2}$
$ω^2 = \frac{[-g[m_1l_1+m_2l_2cos(C)]cos(θ_1^0)-m_2l_2sin(C)sin(θ_1^0)]}{m_1l_1^2+m_2l_2^2}$

$C = arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})=arcsin(\frac{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}{-2l_1l_2})$

now plugging in for $C$; $α(t) = \frac{g[[m_1l_1+m_2l_2[\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}]]sin(θ_1^0)-m_2l_2[\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{-2l_1l_2}]cos(θ_1^0)}{m_1l_1^2+m_2l_2^2}$

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Things look good to me all the way through equation (7). Then, I think you should write ##\theta_1(t) = \theta_1^0 + \epsilon(t)## where ##\theta_1^0## is a constant representing the equilibrium position for ##\theta_1##. Starting with equation (8), everywhere you have ##\theta_1## you should write ##\theta_1^0##.

You can solve for ##\theta_1^0## in terms of ##C## by using equation (7) when the system is hanging in equilibrium so that ##\ddot \theta_1 = 0##.

Near the end you have introduced ##\alpha##. Show that ##\alpha = 0##.

Wavefunction said:
Wavefunction said:

## Homework Statement

Two masses $m_1$ and $m_2 (m_1 ≠ m_2)$ are connected by a rigid rod of length $d$ and of negligible
mass. An extensionless string of length$l_1$ is attached to $m_1$ and connected to a fixed point $P$.
Similarly, a string of length $l_2 (l_1 ≠ l_2)$ connects $m_2$ and $P$. Obtain the equation describing the
motion in the plane of $m_1$, $m_2$, and $P$ and find the frequency of small oscillations around the
equilibrium position.

## Homework Equations

(1) $\frac{∂L}{∂q_j}-\frac{d}{dt}\frac{∂L}{∂\dot{q_j}}+\sum_kλ_k(t)\frac{∂f_k(q_j,t)}{∂q_j} = 0$

(2) $s=Dn-m$

## The Attempt at a Solution

Okay so the first thing I did in this problem was I tried to figure out how many coordinates I need. Each particle can move in 2 dimensions so $D=2$. There are 2 particles so $n=2$. There are 3 constraints:(a) $x_1^2+y_1^2 = l_1^2$, (b) $x_2^2+y_2^2 = l_1^2$, and (c) $(x_2-x_1)^2+(y_2-y_1)^2 = d^2$

$s=(2*2)-3 = 1$ However, I'll start of with the original four coordinates and work my way to only 1:

The Lagrangian:
$L=T-U → T=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]$ and $U=g[m_1y_1+m_2y_2]$
Then $L=\frac{m_1}{2}[\dot{x_1}^2+\dot{y_1}^2]+\frac{m_2}{2}[\dot{x_2}^2+\dot{y_2}^2]-g[m_1y_1+m_2y_2]$

Now I can apply 2 out of my 3 constraints (a and b):
$x_1 = l_1sin(θ_1)$
$y_1 = l_1cos(θ_1)$
$x_2 = l_2sin(θ_2)$
$y_2 = l_2cos(θ_2)$

Applying these transformations to the Lagrangian:

$L=\frac{1}{2}[m_1l_1^2\dot{θ_1}^2+m_2l_2^2\dot{θ_2}^2]-g[m_1l_1cos(θ_1)+m_2l_2cos(θ_2)]$

Now applying the transformations to constraint (c) $l_1^2+l_2^2 -2l_1l_2cos(θ_1-θ_2)=d^2$ This will be my constraint in (1)

Okay now I can use eq. (1):

$θ_1$: $\frac{∂L}{∂θ_1} = g[m_1l_1sin(θ_1)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_1}}] = \frac{d}{dt}[[m_1l_1^2]\dot{θ_1}]$, and $λ\frac{∂f}{∂θ_1} = λ(2l_1l_2sin(θ_1-θ_2))$

(3) $g[m_1l_1sin(θ_1)]-[m_1l_1^2]\ddot{θ_1}+λ(2l_1l_2sin(θ_1-θ_2))=0$

$θ_2$: $\frac{∂L}{∂θ_2} = g[m_2l_2sin(θ_2)]$, $\frac{d}{dt}[\frac{∂L}{∂\dot{θ_2}}] = \frac{d}{dt}[[m_2l_2^2]\dot{θ_2}]$, and $λ\frac{∂f}{∂θ_1} = -λ(2l_1l_2sin(θ_1-θ_2))$

(4) $g[m_2l_2sin(θ_2)]-[m_2l_2^2]\ddot{θ_2}-λ(2l_1l_2sin(θ_1-θ_2))=0$

Okay now I can use the fact that $\dot{θ}$ is the same for both particles:

$\dot{θ_1}=\dot{θ_2}$, $\ddot{θ_1}=\ddot{θ_2}$, and $θ_1= θ_2+C$

Then (3) and (4) become:

(5)$g[m_1l_1sin(θ_1)]-[m_1l_1^2]\ddot{θ_1}+λ(2l_1l_2sin(C))=0$

(6)$g[m_2l_2sin(θ_1-C)]-[m_2l_2^2]\ddot{θ_1}-λ(2l_1l_2sin(C))=0$

Now adding eqs (5) and (6) together:

$g[m_1l_1sin(θ_1)+m_2l_2sin(θ_1-C)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

$g[m_1l_1sin(θ_1)+m_2l_2[sin(θ_1)cos(C)-sin(C)cos(θ_1)]]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

(7) $g[[m_1l_1+m_2l_2cos(C)]sin(θ_1)-m_2l_2sin(C)cos(θ_1)]-[m_1l_1^2+m_2l_2^2]\ddot{θ_1} = 0$

Now if the pendulum is in static equilibrium then $θ_1^0$ is a constant which implies $\ddot{θ_1^0} = 0$. Now if I consider some small angular perturbation $ε(t)$ about $θ_1$ Also $θ_1(t)= θ_1^0+ε(t)$ Solving for $θ_1^0$ in terms of $C$:

$\frac{sin(θ_1^0)}{cos(θ_1^0)}=\frac{m_2l_2sin(C)}{gm_1l_1+m_2l_2cos(C)}$

$θ_1^0= arctan(\frac{m_2l_2sin(C)}{gm_1l_1+m_2l_2cos(C)})$

then eq (7) becomes:

(8) $g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0+ε)-m_2l_2sin(C)cos(θ_1^0+ε)]-[m_1l_1^2+m_2l_2^2][\ddot{θ_1^0}+\ddot{ε}] = 0$

$g[[m_1l_1+m_2l_2cos(C)][sin(ε)cos(θ_1^0)+sin(θ_1^0)cos(ε)]-m_2l_2sin(C)[cos(θ_1^0)cos(ε)-sin(θ_1^0)sin(ε)]]-[m_1l_1^2+m_2l_2^2]\ddot{ε} = 0$

Now since the oscillations are small $sin(ε)≈ε$ and $cos(ε)≈1$

$g[[m_1l_1+m_2l_2cos(C)][εcos(θ_1^0)+sin(θ_1^0)]-m_2l_2sin(C)[cos(θ_1^0)-sin(θ_1^0)ε]]-[m_1l_1^2+m_2l_2^2]\ddot{ε} = 0$

$g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0)-m_2l_2sin(C)cos(θ_1^0) = [m_1l_1^2+m_2l_2^2]\ddot{ε}+[-g[[m_1l_1+m_2l_2cos(C)]cos(θ_1^0)-m_2l_2sin(C)sin(θ_1^0)]ε$

$\ddot{ε}+ω^2ε=α(t)$ where $α(t) = \frac{g[[m_1l_1+m_2l_2cos(C)]sin(θ_1^0)-m_2l_2sin(C)cos(θ_1^0)}{m_1l_1^2+m_2l_2^2}$
$ω^2 = \frac{[-g[m_1l_1+m_2l_2cos(C)]cos(θ_1^0)-m_2l_2sin(C)sin(θ_1^0)]}{m_1l_1^2+m_2l_2^2}$

$C = arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2})=arcsin(\frac{\sqrt{(-2l_1l_2)^2-(d^2-l_1^2-l_2^2)^2}}{-2l_1l_2})$

Notice $α(t) = 0$ because the numerator of $α$ is just the static equilibrium condition applied to eq (7)

Therefore $\ddot{ε}+ω^2ε = 0$

Finally, I can solve for $ω^2$ because I have $C$ and $θ_1^0$

Alright I think I have it down now.

Looks good. You can show that your expression for ##\omega## reduces to what you expect if you treat the system as a physical pendulum.

## 1. What is a double pendulum?

A double pendulum is a physical system consisting of two pendulums attached to each other, where the motion of one pendulum affects the motion of the other.

## 2. What is the Lagrangian method?

The Lagrangian method is a mathematical approach used in classical mechanics to describe the motion of a system by considering its potential and kinetic energy.

## 3. Why is the Lagrangian method used for solving the equations of motion for a double pendulum?

The Lagrangian method allows us to derive a set of equations that describe the motion of a system from a single equation, making it a more efficient and elegant approach than using Newton's laws of motion.

## 4. What are the variables needed to solve for the equations of motion for a double pendulum using the Lagrangian method?

The variables needed include the mass and length of each pendulum, the initial positions and velocities of the pendulums, and the gravitational acceleration.

## 5. Are there any limitations to using the Lagrangian method for solving the equations of motion for a double pendulum?

One limitation is that the Lagrangian method assumes the system is conservative, meaning there is no energy loss due to friction or other external forces. It also assumes the pendulums are rigid bodies, which may not be the case in some real-world scenarios.

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