Two pendulums connected with a massless rope

[LagrangeEuler]

Homework Statement

Two pendulums of same mass and length that oscillate in same horizontal plane are connected with maseless horizontal rope. What is dependence of amplitude of pendulums as a function of time?

Homework Equations

For harmonic oscilation
$$x=x_0\sin(\omega t+\varphi_0)$$
Kinetic energy
$$E_k=\frac{1}{2}mv^2$$
Potential energy
$$E_p=\frac{1}{2}kx^2$$

The Attempt at a Solution

In case from the problem I suppose that kinetic energy is simple
$$E_k=E_{k1}+E_{k2}=\frac{m}{2}(\dot{\varphi}_1^2+\dot{\varphi}_2^2)l^2$$
and potential energy is
$$E_p=E_{p1}+E_{p2}=mgl(1-\cos\varphi_1)+mgl(1-\cos \varphi_2)$$
However I am not sure how from this to get amplitude dependence of pendulums as a function of time.

 

[Apashanka]

Homework Statement

Two pendulums of same mass and length that oscillate in same horizontal plane are connected with maseless horizontal rope. What is dependence of amplitude of pendulums as a function of time?

Homework Equations

For harmonic oscilation
$$x=x_0\sin(\omega t+\varphi_0)$$
Kinetic energy
$$E_k=\frac{1}{2}mv^2$$
Potential energy
$$E_p=\frac{1}{2}kx^2$$

The Attempt at a Solution

In case from the problem I suppose that kinetic energy is simple
$$E_k=E_{k1}+E_{k2}=\frac{m}{2}(\dot{\varphi}_1^2+\dot{\varphi}_2^2)l^2$$
and potential energy is
$$E_p=E_{p1}+E_{p2}=mgl(1-\cos\varphi_1)+mgl(1-\cos \varphi_2)$$
However I am not sure how from this to get amplitude dependence of pendulums as a function of time.[/B]

The lagrangian is KE=ml²(θ₁dot²+θ₂dot²)/2
PE=mgl(θ₁²+θ₂²)/2+Constt.for small angle approximation
Solving this for θ₁ and θ₂ will yield the same equation for a SHM with angular frequency ω²=g/l
And the angular displacement would be θ₁~Asin(ωt) and θ₂~ Bsin(ωt)

 

[Celeritas]

The lagrangian is KE=ml²(θ₁dot²+θ₂dot²)/2
PE=mgl(θ₁²+θ₂²)/2+Constt.for small angle approximation
Solving this for θ₁ and θ₂ will yield the same equation for a SHM with angular frequency ω²=g/l
And the angular displacement would be θ₁~Asin(ωt) and θ₂~ Bsin(ωt)

If they are connected by a massless rope, then surely the dynamics reduces to a 1 degree of freedom system. $\theta_1=\theta_2$.

 

[LagrangeEuler]

It is not that simple I think. Rope has constant length. So what is happening if pendulums go to opposite directions?