Equations of motion two springs pendulum system

[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.