Equations of motion two springs pendulum system

[Springer]

So for the lower part, where the mass is located and you have small displacements in y=vertical en x=horizontal direction, the equations become:

my" = m*g*y*cos(theta) ; as this is the vertical component
mx" = m*g*x*cos(theta) ; as this is het horizontal component

Is this right? Furthermore, how can I combine this upper and lower part to bring the amount of equations down to two for the whole system?

 

[haruspex]

So for the lower part, where the mass is located and you have small displacements in y=vertical en x=horizontal direction, the equations become:

my" = m*g*y*cos(theta) ; as this is the vertical component
mx" = m*g*x*cos(theta) ; as this is het horizontal component

Is this right? Furthermore, how can I combine this upper and lower part to bring the amount of equations down to two for the whole system?

Why the x and y factors? There is no spring here. Besides, mgy would not have the right dimension for a force.
One of the trig functions must be sine, surely?
What connects the mass to the springs?

 

[Springer]

Well, the massless rod connects the mass to the springs. The rod is depending on the position of the springs (x and y direction) so the changing parameter is theta which is the angle between the vertical (no movement) and the actual position of the mass at certain distance L, right?

The general equation for a pendulum is: theta"+(g/L)*sin(theta)=0

When using a mass, you will get:
m*(theta)"+(mg/L)*sin(theta) = 0

Is this the right line of reasoning and how to continue with this?

 

[haruspex]

Well, the massless rod connects the mass to the springs. The rod is depending on the position of the springs (x and y direction) so the changing parameter is theta which is the angle between the vertical (no movement) and the actual position of the mass at certain distance L, right?

The general equation for a pendulum is: theta"+(g/L)*sin(theta)=0

When using a mass, you will get:
m*(theta)"+(mg/L)*sin(theta) = 0

Is this the right line of reasoning and how to continue with this?

Because the top of the rod is moving, I would regard it as unsafe to assume a result from simple pendulums. There may be some subtlety missed. Better to work from first principles, but then apply, as necessary, the same mathematical methods used to obtain the pendulum result.
Yes, the rod connects the mass to the springs, so it is through the tension in the rod that the springs affect the mass. But you left this out of your equations in post #31.
As I mentioned in post #32, there are two other errors in those equations. We need to concentrate on getting those equations right. Reread my post #31 and see if you can correct the equations.

 

[Springer]

Okay so when we consider the balance of the lower part (rod and mass), you get:

Vertical: m*g*cos(theta) - T*cos(theta) = 0
Horizontal: m*g*sin(theta) - T*sin(theta) = 0

Is that correct?

 

[haruspex]

Okay so when we consider the balance of the lower part (rod and mass), you get:

Vertical: m*g*cos(theta) - T*cos(theta) = 0
Horizontal: m*g*sin(theta) - T*sin(theta) = 0

Is that correct?

Gravity is no longer vertical?

 

[Springer]

For the vertical, you are right indeed:
Vertical component: m*g - T*cos(theta) = 0

However, for the horizontal part, you still have a component due to angle theta so that one stays
Horizontal: m*g*sin(theta) - T*sin(theta)

Right?

 

[haruspex]

For the vertical, you are right indeed:
Vertical component: m*g - T*cos(theta) = 0

However, for the horizontal part, you still have a component due to angle theta so that one stays
Horizontal: m*g*sin(theta) - T*sin(theta)

Right?

No. How does gravity get a horizontal component? These are the forces acting, independently, on the mass. The force of gravity knows nothing about the rod, and certainly not its angle.

 

[Springer]

But when you consider only the mass itself, you will say that you have Fz and T under angle theta which gives the following:

m*g + T*cos(theta) = 0

 

[haruspex]

But when you consider only the mass itself, you will say that you have Fz and T under angle theta which gives the following:

m*g + T*cos(theta) = 0

Edit:
What you had for the vertical was fine, mg-Tcos(theta). Whether it is that or with the opposite sign depends on what conventions you adopt. What you have wrong in post #37 is for the horizontal. Gravity is a vertical force by definition of vertical. It has no horizontal component.

More edits: it cannot be mg+Tcos(theta)=0. That leaves out the acceleration of the mass.

 

[haruspex]

Sorry, I may have confused you with my last few posts. When I looked at your post #35 I homed in on errors related to the forces. I completely missed that the mass x acceleration terms were absent.

 

[Springer]

So that results in:

Vertical: m*g - T*cos(theta) = 0
Horizontal: X - T*sin(theta) = 0

But what should X be then? 0

 

[haruspex]

So that results in:

Vertical: m*g - T*cos(theta) = 0
Horizontal: X - T*sin(theta) = 0

But what should X be then? 0

Yes, but read my post #41, which may have crossed with yours.

 

[Springer]

Yes that makes it hard for me. At the moment we have four equations, two for the upper part and two for the lower part

However, in the end I should get two governing equations like:

mu" + ku = F(t) in order to be able to calculate the natural frequencies.

At the moment, I'm a bit confused how to combine them to the two governing equations

 

[haruspex]

Yes that makes it hard for me. At the moment we have four equations, two for the upper part and two for the lower part

However, in the end I should get two governing equations like:

mu" + ku = F(t) in order to be able to calculate the natural frequencies.

At the moment, I'm a bit confused how to combine them to the two governing equations

To make sure we're still on track, please post those four equations as you now have them.

 

[Springer]

Upper part:

The vertical y => 0 = -2*k*0,5*y*sqrt(2) + T*cos(theta)
The horizontal x => 0 = -0,5*k*sqrt(2)*(0+x) + 0,5*k*sqrt(2)*(0-x) + T*sin(theta)

Lower part:
Vertical: m*g - T*cos(theta) = 0
Horizontal: X - T*sin(theta) = 0

 

[haruspex]

Upper part:

The vertical y => 0 = -2*k*0,5*y*sqrt(2) + T*cos(theta)
The horizontal x => 0 = -0,5*k*sqrt(2)*(0+x) + 0,5*k*sqrt(2)*(0-x) + T*sin(theta)

Lower part:
Vertical: m*g - T*cos(theta) = 0
Horizontal: X - T*sin(theta) = 0

You are still not taking action on what I told you in posts 40 and 41.
I finally got you to stop putting mass x acceleration terms in tne upper part equation, there being no mass there, but in the lower part there is mass and it accelerates. So in the lower part there is a net force, not 0.

When the junction is displaced (x,y) from equilibrium position, and the rod is displaced a small angle theta anticlockwise from the vertical, what is the displacement of the mass from its equilibrium position?

 

[Springer]

Yes but in the equations of the upper part, i did not include mass, only the reaction force T of the rod

For the lower part, I get
Vertical: mass*acceleration = m*g - T*cos(theta) ?
Horizontal: mass*acceleration = X - T*sin(theta) ?

 

[haruspex]

Yes but in the equations of the upper part, i did not include mass, only the reaction force T of the rod

For the lower part, I get
Vertical: mass*acceleration = m*g - T*cos(theta) ?
Horizontal: mass*acceleration = X - T*sin(theta) ?

Right, except you can throw away the X. (Why is it there? What do you think it represents?)
You can apply small angle approximations to the trig functions.

The challenge now is to write expressions for the horizontal and vertical accelerations of the mass in terms of x, y and theta, where (x,y) is the displacement of the junction. As a first step, express the displacement of the mass in terms of those three variables.