Lagrangian of a mass bewteen two springs with a pendulum hanging down

[Davidllerenav]

Homework Statement

A particle of mass $m_1$ hangs from a rod of negligible mass and length $l$, whose support point consists of another particle of mass $m_2$ that moves horizontally subject to two springs of constant $k$ each one. Find the equations of motion for this system.

Relevant Equations

$L=T-V$

What I first did was setting the reference system on the left corner. Then, I said that the position of the mass $m_2$ is $x_2$. I also supposed that the pendulum makes an angle $\theta$ with respect to the vertical axis $y$. So the generalized coordinates of the system would be $x_2$ and $\theta$. Thus, the coordinates of $m_1$ are:

• $x_1=x_2+l\sin\theta$
• $y_1=-l\cos\theta$

Then I took the time derivative of both $x_1$ and $y_1$:

• $\dot x_1=\dot x_2+l\dot\theta\cos\theta$
• $\dot y_1=l\dot\theta\sin\theta$

The Kinetic energy $T$ would be the sum of the kinetic energy of each mass $T=T_1+T_2$. Since $T1=\frac{1}{2}m_2\dotx_2^2$ and $T_2=\frac{1}{2}m_1(\dot x_1^2+\dot y_1^2$, I got:

• $T=\frac{1}{2}m_2\dot x_2^2+\frac{1}{2}m_1(\dot x_2^2+l^2\dot\theta^2+2\dot x_2\dot\theta l)$

Now, I have trouble with the potential energy. The potential energy of $m_1$ is easy, it would be just $V_1=m_1gh$:

• $V_1=-m_1gl\cos\theta$

The potential energy of $m_2$ would be the sum of the potential energy that each spring has on the mass. What I did was saying that that the mass $m_2$ moves a distance $x_2$, thus the lhs spring would have a displacement of $x_2$ and the rhs spring would have a contraction of $x_2$ likewise. So:

• $V_2=\frac{1}{2}k x_2^2+\frac{1}{2}k(-x_2)^2=kx_1^2$

But, according to my teacher, if the first spring moves a distance $x_2$, then the second would contract a distance of $a-x_2$, where $a$ is the distance between the walls, that is constant. I don't understand why this is the case, can you please explain it better for me to understand? Thanks.

 

[anuttarasammyak]

Say
$$x_2=\frac{a}{2}$$ , natual length of spring,
$$V_2=0$$
So
$$V_2=k(x_2-\frac{a}{2})^2$$

 

[Marc Rindermann]

I don't understand it either. If $s_1$ and $s_2$ are the lengths of both springs then $a = s_1 + s_2$.

Now, if one spring stretches by $x$ then the other spring must be compressed by the same amount, $(s_1 + x) + (s_2 - x) = a$

And if $V_1 = \frac12 k x^2$ then $V_2 = \frac12 k x^2$ and so $V = V_1 + V_2 = k x^2$

Although, I have the origin where $m_2$ is at rest. You say your origin is in the left corner. So the left spring is attached to the origin. Your teacher attaches the the right spring to $a$.

Now if you move $m_2$ the new location will be at $x_2$ as measured from the origin for the left spring, and at $a - x_2$ which is measured from the right wall for the right spring.

 

[anuttarasammyak]

What I first did was setting the reference system on the left corner.

x＝0 at the left corner. Do you count V2=0 for x2=0,i.e. the mass is on the left corner ?

 

[Davidllerenav]

x＝0 at the left corner. Do you count V2=0 for x2=0,i.e. the mass is on the left corner ?

No. I count $V_2=0$ when $x_2=a/2$, i.e. when the mass is halfway between the two walls.

 

[Davidllerenav]

I don't understand it either. If $s_1$ and $s_2$ are the lengths of both springs then $a = s_1 + s_2$.

Now, if one spring stretches by $x$ then the other spring must be compressed by the same amount, $(s_1 + x) + (s_2 - x) = a$

And if $V_1 = \frac12 k x^2$ then $V_2 = \frac12 k x^2$ and so $V = V_1 + V_2 = k x^2$

Although, I have the origin where $m_2$ is at rest. You say your origin is in the left corner. So the left spring is attached to the origin. Your teacher attaches the the right spring to $a$.

Now if you move $m_2$ the new location will be at $x_2$ as measured from the origin for the left spring, and at $a - x_2$ which is measured from the right wall for the right spring.

So, the result I got would only be correct if the origin is halfway between the two walls?

 

[anuttarasammyak]

Yes. And for your original coordinate see #2.

 

[Davidllerenav]

Say
$$x_2=\frac{a}{2}$$ , natual length of spring,
$$V_2=0$$
So
$$V_2=k(x_2-\frac{a}{2})^2$$

I'm I bit confused by the use of $x_2$, so let's say that $x_2$ is the position of the mass at its equilibrium position, i.e. $x_2=a/2$ as you said. And let $x$ be the position of the mass at some time Then if the mass moves to the right to a point $x$, the left spring would stretch a distance of $x-x_2=x-a/2$. But, wouldn't the spring on the right be compressed by the same amount?

 

[anuttarasammyak]

That's the reason why $k$ ,not $\frac{k}{2}$ for a single spring, in $V_2$. Doubled as you did yourself in #1 and #3.

 

[kuruman]

Homework Statement:: A particle of mass $m_1$ hangs from a rod of negligible mass and length $l$, whose support point consists of another particle of mass $m_2$ that moves horizontally subject to two springs of constant $k$ each one. Find the equations of motion for this system.
Relevant Equations:: $L=T-V$

But, according to my teacher, if the first spring moves a distance $x_2$ , then the second would contract a distance of $a-x_2$, where is the distance between the walls, that is constant. I don't understand why this is the case, can you please explain it better for me to understand?

Your generalized coordinates are relative to an origin that is at $m_2$ when the system is at equilibrium, string vertical, springs relaxed. This choice allows you to use the same symbol $x_2$ to denote both the generalized coordinate and the amount by which each spring is compressed or stretched. Thus, if $m_2$ is displaced to the right, the elastic potential energy is as you have written it $V_{\text{el}}=\frac{1}{2}kx_2^2+\frac{1}{2}kx_2^2=kx_2^2.$

Now suppose you choose to measure generalized coordinate $x_2$ relative to the left wall. If $m_2$ is again displaced from equilibrium to the right, the amount by which the left spring is stretched is $x_2-\frac{a}{2}$ and the amount by which the right spring is compressed is $\frac{a}{2}-x_2$ and the potential energy is $V_{\text{el}}=\frac{1}{2}k(x_2-\frac{a}{2})^2+\frac{1}{2}k(\frac{a}{2}-x_2)^2=k(x_2-\frac{a}{2})^2.$

In such problems involving springs and Lagrangians one needs to be careful to enunciate the differences, if any, between generalized coordinates and spring displacements.