Two pendulums connected with a massless rope

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SUMMARY

The discussion focuses on the dynamics of two pendulums of equal mass and length, connected by a massless rope, and their amplitude dependence as a function of time. The kinetic energy is expressed as E_k = (1/2)m(θ̇₁² + θ̇₂²)l², while the potential energy is given by E_p = mg(l(1 - cos(θ₁)) + l(1 - cos(θ₂))). The solution involves using the Lagrangian approach, leading to the conclusion that the system behaves like a simple harmonic oscillator with angular frequency ω² = g/l. The angular displacements are represented as θ₁ ~ A sin(ωt) and θ₂ ~ B sin(ωt), indicating a coupled motion.

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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.
 
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LagrangeEuler said:

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=ml21dot22dot2)/2
PE=mgl(θ1222)/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)
 
Apashanka said:
The lagrangian is KE=ml21dot22dot2)/2
PE=mgl(θ1222)/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##.
 
It is not that simple I think. Rope has constant length. So what is happening if pendulums go to opposite directions?
 

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