- #1

Wavefunction

- 99

- 4

## 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(: