Two pendulums connected with a massless rope

  • Thread starter Thread starter LagrangeEuler
  • Start date Start date
  • Tags Tags
    Massless Rope
LagrangeEuler
Messages
711
Reaction score
22

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.
 
Last edited by a moderator:
Physics news on Phys.org
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?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top