# Equations of motion two springs pendulum system

• Springer
In summary, the conversation discusses the governing equations for a system in static equilibrium with small vibrations. The system consists of two springs, a massless rod, and a point mass attached to immovable supports. The equations discussed are based on the displacement method and Newton's equations, with the system having three degrees of freedom. The conversation also mentions the need to consider the angle of the rod as a variable in the equations.
Springer

## Homework Statement

Give governing equations for the system about its static equilibrium, assuming small vibrations

System consists of two springs located under 45 degrees to the vertical (both have same k-value) in undisturbed situation. Lower ends of the springs are attached to each other and to massless rod with length D, upper ends to immovable supports. The massless rod has point mass (m) at the lower end. System can move vertically and horizontally and is exposed to gravity.

## Homework Equations

Displacement method: mx"+cx'+kx = F(t)

## The Attempt at a Solution

System has 3 degrees of freedom (horizontal (x2) , vertical (x1) and rotation of pendulum around supension (theta))

I suggested:
mx"1 = -1/2*(k*sqrt(2)*x1) - 1/2*(k*sqrt(2)*x1) + mg
mx"2 = -1/2*(k*sqrt(2)*x2) + 1/2*(k*sqrt(2)*x2)
J(theta)" = -m*g*D*(theta)

The last equation under the assumption of small angles of the massless rod.

However, in this case it results in a decoupled dynamic system which I suppose is incorrect?

Thanks in advance for all support!

Can you draw it?
I can't see what system you are looking at, I have some ideas but guessing is not efficient.

Are you using the Newton equations or the Euler-Lagrange equations? (I know which I would use)

It is mentioned to use the Newton equations. The schematization is like this:

Springer said:
It is mentioned to use the Newton equations. The schematization is like this:

View attachment 85498
Are your x1 and x2 displacements of the mass or displacements of the Y junction?

I assumed displacement of the Y junction as the mass is connected (fixed connection) to the rod.

Springer said:
I assumed displacement of the Y junction as the mass is connected (fixed connection) to the rod.
Your first two equations treat the mass as being at the Y junction. You need to allow that the rod is not always vertical.

That was the reason I used the third equation for small angles of the rod to it's static (vertical) equilibrium. How do you suppose I should link the horizontal and vertical displacement of the whole system with the mass on certain distance, expressed in the two first equations?

Springer said:
That was the reason I used the third equation for small angles of the rod to it's static (vertical) equilibrium. How do you suppose I should link the horizontal and vertical displacement of the whole system with the mass on certain distance, expressed in the two first equations?
Adding an equation doesn't correct a wrong equation. Let the tension in the rod be T and consider the FBDs for the junction and the mass separately.

As suggested, you would consider the body diagram as schematized in the figure above as starting point for this issue, Haruspex? For the upper part, there will be two equations for displacement in horizontal direction and vertical direction due to Force T and for the lower part a reaction Force R and gravity Force=m*g?

Springer said:
View attachment 87245

As suggested, you would consider the body diagram as schematized in the figure above as starting point for this issue, Haruspex? For the upper part, there will be two equations for displacement in horizontal direction and vertical direction due to Force T and for the lower part a reaction Force R and gravity Force=m*g?
Ok, but you should show the angle of the rod as a variable too.

Dear Haruspex,

As you proposed, I had the same idea as you can see in the attachment. The angle between the rod and the vertical is indicated as theta. In this case, should you get two governing equations, one for the upper part including theta1 and one for the lower part including theta2 where theta1=theta2?

#### Attachments

• Screen Shot 2015-12-09 at 10.31.44.png
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Springer said:
Dear Haruspex,

As you proposed, I had the same idea as you can see in the attachment. The angle between the rod and the vertical is indicated as theta. In this case, should you get two governing equations, one for the upper part including theta1 and one for the lower part including theta2 where theta1=theta2?
Yes.

Hi Haruspex,

When I make the FBD of the upper part (1) and lower part (2), I get the following equations (where u=displacement in vertical direction):

1) mu" = -2*(k*sqrt2*u)/2 + T; T = m*g*sin(theta)
2) mu" = -R + T; R = reaction force, T = m*g*sin(theta)

However, the displacements in horizontal direction are not incorporated in the equations, which is not correct right?
Thank you for your help!

Springer said:
Hi Haruspex,

When I make the FBD of the upper part (1) and lower part (2), I get the following equations (where u=displacement in vertical direction):

1) mu" = -2*(k*sqrt2*u)/2 + T; T = m*g*sin(theta)
2) mu" = -R + T; R = reaction force, T = m*g*sin(theta)

However, the displacements in horizontal direction are not incorporated in the equations, which is not correct right?
Thank you for your help!
Your vertical equations are incorrect.
T is not equal to m g sin(theta), in general. The FBD for for the junction doesn't 'know' about the mass. It just feels the tension in the rod. So the left part of (1) is right, but throw away the second part.
The rod is massless, so. What is the relationship between T and R?
Likewise, in (2), you don't want both T and R. All the mass knows about are gravity, the pull from the rod, and its own acceleration.

I reconsidered the equations as you suggested. The relationship between T and R is that they have to make a force balance; in other words T = -R. With this input, I get the following for the upper part (1) and lower part (2):

1) mu" = -2*(k*sqrt2*u)/2 + T
2) mu" = m*g*sin(theta) - R

Is this what you meant, where you can balance the forces of T and R?

Springer said:
I reconsidered the equations as you suggested. The relationship between T and R is that they have to make a force balance; in other words T = -R. With this input, I get the following for the upper part (1) and lower part (2):

1) mu" = -2*(k*sqrt2*u)/2 + T
2) mu" = m*g*sin(theta) - R

Is this what you meant, where you can balance the forces of T and R?
I'm sorry, I made a mistake in my previous post. The left hand side of your (1) was not right either.

There is no mass at the junction, so forces there must completely balance.
We must consider an x displacement and a y displacement of the junction. What is the sum of forces that results in each (x and y) direction? Each of these sums will be zero.

You can express the displacement of the mass in terms of the above x and y and the angle theta. What are the vertical and horizontal sums of forces there?

Okay so when I just consider the junction, you get the following:

For displacement y=vertical displacement
my" = -2*k* 0,5*y*sqrt(2)

For displacement x=horizontal displacement
mx" = -k* 0,5*x*sqrt(2)

In this consideration, the change in angle of 45 degrees of the springs to their support is assumed as small. Is this reasoning correct?

Springer said:
Okay so when I just consider the junction, you get the following:

For displacement y=vertical displacement
my" = -2*k* 0,5*y*sqrt(2)

For displacement x=horizontal displacement
mx" = -k* 0,5*x*sqrt(2)

In this consideration, the change in angle of 45 degrees of the springs to their support is assumed as small. Is this reasoning correct?
There is no mass at the junction. Instead, there is a force from the rod, which you have now omitted. You had that previously.

Other than that, the y equation is right - it's not exact, but it is a valid approximation for small displacements. The exact form would use both x and y displacements to find the spring extension, then find the vertical component of the resulting force. The approximation is ok because the x displacement increases the vertical component of one tension and reduces the vertical component of the orher in roughly equal amounts.
The same arguments apply in the horizontal direction, but this time you did not get quite the right expression for the force. There are two springs.

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 extends and the right spring shortens, right?

I am sorry, indeed I omitted T which is the force from the rod. So the y=vertical displacement gives the following equation:
my" = -2*k*0,5*y*sqrt(2) + T

These equations are for the upper part but how do I combine these with the lower part (where the mass is located)? In y=vertical direction, the only vertical forces are m*g and reaction force R as far I know. In x=horizontal direction, due to the angle theta that results from the displacement of the junction, the horizontal forces of the mass are R*cos(theta) right?

Thanks a lot.

Springer said:
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 extends and the right spring shortens, right?

I am sorry, indeed I omitted T which is the force from the rod. So the y=vertical displacement gives the following equation:
my" = -2*k*0,5*y*sqrt(2) + T

These equations are for the upper part but how do I combine these with the lower part (where the mass is located)? In y=vertical direction, the only vertical forces are m*g and reaction force R as far I know. In x=horizontal direction, due to the angle theta that results from the displacement of the junction, the horizontal forces of the mass are R*cos(theta) right?

Thanks a lot.
You don't seem to be grasping the significance of the fact that there is no mass at the junction. It means there is no mass to be associated with the acceleration. The applied forces there, two springs and rod, must exactly balance. ##\Sigma F=0##.
Also, you have left out the rod's tension in the X equation.
I don't know you introduced kleft and kright for horizontal but use k for both in the verical. Each spring only has one constant, and the springs are the same,

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 the reaction forces in the springs. Is that incorrect? In other words; as there is no mass associated with the acceleration, there is no force from the rod?

Another question, the exercise mentions that the lower ends of the springs are connected to each other and to a rigid massless rod. In which way can I incorporate the rod's tension in horizontal direction (which is x here)? I assumed that the rod (with length L but massless) is just a tool to get inertia as it is at certain distance from the junction.

Springer said:
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 the reaction forces in the springs. Is that incorrect? In other words; as there is no mass associated with the acceleration, there is no force from the rod?

Another question, the exercise mentions that the lower ends of the springs are connected to each other and to a rigid massless rod. In which way can I incorporate the rod's tension in horizontal direction (which is x here)? I assumed that the rod (with length L but massless) is just a tool to get inertia as it is at certain distance from the junction.
Focus on the junction. There are three forces acting on it, the tensions in the two springs and the tension in the rod. The mass itself is not in this FBD. Since there is no mass, these three forces must be in balance, and it is unnecessary to consider the acceleration of the junction.
Since we are only considering small displacements, we can take the spring tensions as always acting at 45 degrees. The rod tension acts at varying angle theta.
You have correctly determined the horizontal and vertical components of the spring tensions. If the tension in the rod is T, you can easily write down the vertical and horizontal components of that. So write down two equations, one for the vertical balance of forces at the junction, the other for the horizontal balance there.

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 horizontal and vertical displacement of the springs, you will get the following:

Vertical balance of forces at junction => y = -2*k*0,5*y*sqrt(2) + T*cos(theta)
Horizontal balance of forces at junction => x = -0,5*k*x*sqrt(2) + 0,5*k*x*sqrt(2) + T*sin(theta)

Based on the principle of SigmaF=0 in the junction

Springer said:
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 horizontal and vertical displacement of the springs, you will get the following:

Vertical balance of forces at junction => y = -2*k*0,5*y*sqrt(2) + T*cos(theta)
Horizontal balance of forces at junction => x = -0,5*k*x*sqrt(2) + 0,5*k*x*sqrt(2) + T*sin(theta)

Based on the principle of SigmaF=0 in the junction
Nearly right, but you have the x components of the spring tensions cancelling. That cannot be right, can it? Bear in mind that both springs will be under tension because of the vertical pull, but in the horizontal direction they will cancel when x=0. What we need to find is the change in the horizontal components of the tensions when x is small nonzero. One will increase, while the other decreases.

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 = -0,5*k*sqrt(2)*(x+delta) + 0,5*k*sqrt(2)*(x-delta)

In this case, the two components will not cancel out each other. However, you will get another variable delta. Do you suppose another way to express this or would you apply the same line of reasoning?

Springer said:
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 = -0,5*k*sqrt(2)*(x+delta) + 0,5*k*sqrt(2)*(x-delta)

In this case, the two components will not cancel out each other. However, you will get another variable delta. Do you suppose another way to express this or would you apply the same line of reasoning?
The variable x is the delta. The equilibrium position is at x=0. Set each x to 0 in your equation and change the deltas to x.

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 to complete these two equations with the forces arising of the lower part (where the mass is located) right?

Springer said:
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 to complete these two equations with the forces arising of the lower part (where the mass is located) right?
The above two equations are right if you replace the left hand sides with zeroes (I thought we'd got past this hurdle).

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: mu"+ku=F(t)

Springer said:
However, we did not include the equations of the mass, right?
That's right, you now need the two equations for the acceleration of the mass and the forces acting on it.

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?

Springer said:
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?

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?

Springer said:
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.

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?

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