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

[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 lengthl_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)]

In addition:
\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<br />

[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<br />

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 &lt; 0 and θ_2&gt;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<br />

[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<br />

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<br />

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}<br />

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}<br />

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(:

 

[maajdl]

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

 

[Wavefunction]

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

 

[maajdl]

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.

 

[Wavefunction]

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(:

 

[TSny]

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.

 

[Wavefunction]

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]

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!

 

[TSny]

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.

 

[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 lengthl_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}

 

[TSny]

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]

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 lengthl_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.

 

[TSny]

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