# What is the Lagrangian, equations of motion for this system?

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Description of the system:
The masses m1 and m2 lie on a smooth surface. The masses are attached with a spring of non stretched length l0 and spring constant k. A constant force F is being applied to m2.

My coordinates:
Left of m1 there is a stationary wall. The distance between m1's left surface and the wall is represented as y and the distance between the right and left surfaces of m1 and m2, respectively, is represented as x.

What my brain says:
The kinetic energy of this system is equal to the sum of kinetic energies of the two masses, i.e,
$$\frac{1}{2}m_1 \dot{y}^2 + \frac{1}{2}m_2 (\dot{x}+\dot{y})^2$$
and the potential energy is just the spring potential, i.e,
$$\frac{1}{2} k (x-l_0)^2$$
Therefore, the Lagrangian seems to be,
$$L=\frac{1}{2}m_1 \dot{y}^2 + \frac{1}{2}m_2 (\dot{x}+\dot{y})^2 - \frac{1}{2} k (x-l_0)^2$$

Now, the generalized force is
$$F\cdot \frac{\partial {(x+y)}}{\partial x}$$
So, the equations of motion seem to be,
$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot x}}) - \frac{\partial L}{\partial { x}}=F$$
and
$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot y}}) - \frac{\partial L}{\partial { y}}=0$$

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Orodruin
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Your coordinates are set up in such a manner that the displacement depends on both x and y.

Also note that you perfectly well can treat the force using a potential, since it is conservative.

Your coordinates are set up in such a manner that the displacement depends on both x and y.
Sir, What is the problem if the displacement of m2 depends on both x and y? How can I modify my equations for that? Is there any problem with my generalized force calculation?

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Delta2
Orodruin
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The generalised force for any coordinate is given by ##F(\partial \ell/dq^i)##, where ##\ell## is the displacement and ##F## the force. You have ignored the effect of the generalised force on the ##y## coordinate.

You have ignored the effect of the generalised force on the yyy coordinate.
Is it not $$F\cdot \frac{\partial {(x+y)}}{\partial x}$$
The factor of F becomes 1. Does it not?

Orodruin
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Is it not $$F\cdot \frac{\partial {(x+y)}}{\partial x}$$
The factor of F becomes 1. Does it not?
That is just the effect on the ##x## coordinate, but you are ignoring the fact that ##x + y## also depends on ##y##.

That is just the effect on the xxx coordinate, but you are ignoring the fact that x+yx+yx + y also depends on yyy.
Then my second equation of motion is $$\frac{d}{dt}(\frac{\partial L}{\partial {\dot y}}) - \frac{\partial L}{\partial { y}}=F\cdot \frac{\partial {(x+y)}}{\partial y}$$? Is it correct now?

Orodruin
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Yes. However, as I said, it is easier to do this by including a potential in the Lagrangian.

Sir, what are the finally correct equations? Are they
$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot x}}) - \frac{\partial L}{\partial { x}}=F$$

and

$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot y}}) - \frac{\partial L}{\partial { y}}=F$$