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

[Wavefunction]

Homework Statement

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

Homework Equations

(1) [itex]\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 [/itex]

(2) [itex] s=Dn-m [/itex]

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 [itex] D=2 [/itex]. There are 2 particles so [itex] n=2 [/itex]. There are 3 constraints:(a) [itex] x_1^2+y_1^2 = l_1^2 [/itex], (b) [itex] x_2^2+y_2^2 = l_1^2 [/itex], and (c) [itex] (x_2-x_1)^2+(y_2-y_1)^2 = d^2 [/itex]

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

The Lagrangian:
[itex] 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] [/itex] and [itex] U=g[m_1y_1+m_2y_2] [/itex]
Then [itex] 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] [/itex]

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

Applying these transformations to the Lagrangian:

[itex] 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)] [/itex]

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

Applying this to L:

[itex] 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)] [/itex]

In addition:
[itex] \dot{θ_1}=\dot{θ_2} → θ_1=θ_2+C → f = θ_1-θ_2-C=0 [/itex]

Okay now I can use eq. (1):

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

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

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

(4) [itex] g[m_2l_2sin(θ_2)]-λ=0 [/itex]

From eq. 4 I know [itex] λ [/itex] in terms of [itex] θ_1 [/itex]:

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

Then eq. 3 becomes:

[itex] 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 [/itex]

[itex] 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
[/itex]

[itex] [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
[/itex]

Now there is the matter of [itex] C [/itex] to handle:

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

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

From the law of cosines I get:

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

[itex] C = arccos(\frac{d^2-l_1^2-l_2^2}{-2l_1l_2}) [/itex]

Now plugging in [itex] C [/itex] into the new form of eq. 3:

[itex] [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
[/itex]

[itex] [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
[/itex]

Okay so now I only want to consider small oscillations about the equilibrium position [itex] \Rightarrow θ_1\ll 1 [/itex]

Then eq. 3 becomes the Linear ODE:

[itex] 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
[/itex]

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

[itex] \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}
[/itex]

Now then [itex] 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} [/itex] (frequency of oscillation about the equilibrium point)

The equilibrium condition is given as [itex] θ_1(0) = \frac{f(t)}{ω_0^2} [/itex] because in equilibrium [itex] \ddot{θ_1(t=0)}=0 [/itex] In this case [itex] 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}
[/itex]

The general solution to eq 3. is:

(4) [itex] θ_1(t) = Acos(ω_0t)+Bsin(ω_0t)+θ_p(t) [/itex]

[itex] \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}) [/itex]

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

[itex] \dot{θ_1} = Bω_0cos(ω_0t) → \dot{θ_1(0)} = Bω_0 → B=\frac{ \dot{θ_1(0)}}{ω_0} [/itex]

So the general solution is then:

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

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 [itex] l_i [/itex] from point
[itex] P [/itex]

 

[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 [itex] l_1= l_2 [/itex] and [itex] m_1 = m_2 [/itex] because in that limit I should be able to calculate the reduced mass [itex] \mu [/itex] 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 [itex] \frac{∂L}{∂\dot{θ_2}} [/itex] because I never got that [itex] \frac{∂L}{∂θ_2}= 0 [/itex] I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations. What about [itex]\dot{θ_1}=\dot{θ_2}[/itex]? 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 [itex] d [/itex] in the plane.

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

 

[Wavefunction]

In 1) were you referring to [itex] \frac{∂L}{∂\dot{θ_2}} [/itex] because I never got that [itex] \frac{∂L}{∂θ_2}= 0 [/itex] I think I'm still ok here though because 2 of my 3 constraints were simple coordinate transformations. What about [itex]\dot{θ_1}=\dot{θ_2}[/itex]? 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 [itex] d [/itex] 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 [itex] ε [/itex] the angular displacement from [itex] θ_1(0) [/itex] is what is small. Thank you for pointing that out kind sir!

 

[TSny]

In 1) were you referring to [itex] \frac{∂L}{∂\dot{θ_2}} [/itex] because I never got that [itex] \frac{∂L}{∂θ_2}= 0 [/itex]

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 [itex]\dot{θ_1}=\dot{θ_2}[/itex]? 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 [itex] m_1 [/itex] and [itex] m_2 (m_1 ≠ m_2) [/itex] are connected by a rigid rod of length [itex] d [/itex] and of negligible
mass. An extensionless string of length[itex] l_1[/itex] is attached to [itex] m_1 [/itex] and connected to a fixed point [itex] P [/itex].
Similarly, a string of length [itex] l_2 (l_1 ≠ l_2) [/itex] connects [itex] m_2 [/itex] and [itex] P [/itex]. Obtain the equation describing the
motion in the plane of [itex] m_1 [/itex], [itex] m_2 [/itex], and [itex] P [/itex] and find the frequency of small oscillations around the
equilibrium position.

Homework Equations

(1) [itex]\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 [/itex]

(2) [itex] s=Dn-m [/itex]

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 [itex] D=2 [/itex]. There are 2 particles so [itex] n=2 [/itex]. There are 3 constraints:(a) [itex] x_1^2+y_1^2 = l_1^2 [/itex], (b) [itex] x_2^2+y_2^2 = l_1^2 [/itex], and (c) [itex] (x_2-x_1)^2+(y_2-y_1)^2 = d^2 [/itex]

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

The Lagrangian:
[itex] 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] [/itex] and [itex] U=g[m_1y_1+m_2y_2] [/itex]
Then [itex] 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] [/itex]

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

Applying these transformations to the Lagrangian:

[itex] 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)] [/itex]

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

Okay now I can use eq. (1):

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

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

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

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

Okay now I can use the fact that [itex] \dot{θ} [/itex] is the same for both particles:

[itex] \dot{θ_1}=\dot{θ_2} [/itex], [itex] \ddot{θ_1}=\ddot{θ_2} [/itex], and [itex] θ_1= θ_2+C [/itex]

Then (3) and (4) become:

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

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

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

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

[itex] 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 [/itex]

(7) [itex] 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 [/itex]

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

(8) [itex] 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 [/itex]

[itex] 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 [/itex]

Now since the oscillations are small [itex] sin(ε)≈ε [/itex] and [itex] cos(ε)≈1 [/itex]

[itex] 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 [/itex]

[itex] 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 [/itex]

[itex] \ddot{ε}+ω^2ε=α(t) [/itex] where [itex] 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} [/itex]
[itex]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}[/itex]

[itex] 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}) [/itex]

now plugging in for [itex] C [/itex]; [itex] 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} [/itex]

 

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

Homework Equations

(1) [itex]\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 [/itex]

(2) [itex] s=Dn-m [/itex]

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 [itex] D=2 [/itex]. There are 2 particles so [itex] n=2 [/itex]. There are 3 constraints:(a) [itex] x_1^2+y_1^2 = l_1^2 [/itex], (b) [itex] x_2^2+y_2^2 = l_1^2 [/itex], and (c) [itex] (x_2-x_1)^2+(y_2-y_1)^2 = d^2 [/itex]

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

The Lagrangian:
[itex] 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] [/itex] and [itex] U=g[m_1y_1+m_2y_2] [/itex]
Then [itex] 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] [/itex]

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

Applying these transformations to the Lagrangian:

[itex] 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)] [/itex]

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

Okay now I can use eq. (1):

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

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

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

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

Okay now I can use the fact that [itex] \dot{θ} [/itex] is the same for both particles:

[itex] \dot{θ_1}=\dot{θ_2} [/itex], [itex] \ddot{θ_1}=\ddot{θ_2} [/itex], and [itex] θ_1= θ_2+C [/itex]

Then (3) and (4) become:

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

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

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

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

[itex] 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 [/itex]

(7) [itex] 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 [/itex]

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

[itex] \frac{sin(θ_1^0)}{cos(θ_1^0)}=\frac{m_2l_2sin(C)}{gm_1l_1+m_2l_2cos(C)}[/itex]

[itex] θ_1^0= arctan(\frac{m_2l_2sin(C)}{gm_1l_1+m_2l_2cos(C)}) [/itex]

then eq (7) becomes:

(8) [itex] 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 [/itex]

[itex] 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 [/itex]

Now since the oscillations are small [itex] sin(ε)≈ε [/itex] and [itex] cos(ε)≈1 [/itex]

[itex] 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 [/itex]

[itex] 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 [/itex]

[itex] \ddot{ε}+ω^2ε=α(t) [/itex] where [itex] 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} [/itex]
[itex]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}[/itex]

[itex] 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}) [/itex]

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

Therefore [itex]\ddot{ε}+ω^2ε = 0[/itex]

Finally, I can solve for [itex]ω^2[/itex] because I have [itex] C[/itex] and [itex]θ_1^0[/itex]

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.