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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) ?

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

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...

So that results in: Vertical: m*g - T*cos(theta) = 0 Horizontal: X - T*sin(theta) = 0 But what should X be then? 0

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

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?

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?

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...

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...

So 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) However, we did not include the equations of the mass, right? Moreover, in the end I should get the governing equations in the order of...

Exactly, that was the missing link. So in the end, you'll get the following two equations for the upper part: For the vertical component: my" = -2*k*0,5*y*sqrt(2) + T*cos(theta) For the horizontal component: mx" = -0,5*k*sqrt(2)*(0+x) + 0,5*k*sqrt(2)*(0-x) + T*sin(theta) So now I have...

Hi Haruspex, That is the question I had as well. So if you consider a small displacement "delta" under the assumption that the angle between the springs and their supports stays the same (45 degrees), you'll get the following equation for the horizontal balance of forces at the junction: x =...

As you said in the previous post, the horizontal and vertical components of the spring tension are correctly determined so focussing on the tension T in the rod gives for: Vertical component => y = T*cos(theta) Horizontal component => x = T*sin(theta) When this is combined with the...

Hi Haruspex, In the previous post, you mentioned that "There is no mass at the junction. Instead, there is a force from the rod, which you have now omitted. You had that previously." For the vertical displacement (which is y here) I considered this "force from the rod" as T, in combination with...

So let's call it the left spring and right spring with corresponding k_left and k_right. For example, applying a displacement of the junction to the right (x = horizontal small displacements) gives than: mx" = -0,5*k_left*x*sqrt(2) + 0,5*k_right*x*sqrt(2) This means that the left spring...