Equation of motion for a massless spring system

[mintsnapple]

Homework Statement

Homework Equations

f_spring = k(x_near - x_far)

The Attempt at a Solution

a. FBD:

The force goes through the nodes, and the sum of the forces must be 0 because the nodes are massless. Therefore, kx_1 = x(t)
So x_1(t) = x(t)/k

b. FBD:

For this system, the parallel springs have a k_eq of k + 2k + 3k = 6k. This is in series with the 5k spring, so the k_eq for both is 1/k_eq = 1/5k + 1/6k
= 5/30k + 6/30k
=11/30k
So k_eq = 30k/11.
Therefore, 30k/11(x_3 – x_1) = x(t)
So x_3(t) – x_1(t) = 11x(t)/(30k)

c. FBD:

Since the k_eq = 2k + 3k +k = 6k,

6k(x_2 – x_1) = x(t)
So x_2(t) – x_1(t) = x(t)/(6k)

Is my way of thinking correct?

 

[Chestermiller]

I don't think your thinking is correct. Please write a force balance on each of the nodes. You will then get a set of linear equations that you can solve for each of the nodal displacements in terms of x(t). You should end up with three equations in three unknowns.

Chet

 

[mintsnapple]

Thank you for the reply. Are my FBDs right though?

So assuming that they are:
For node 1: k(x_1) = x(t)
For node 2: 6k(x_2 – x_1) = x(t)
For node 3: 30k/11(x_3 – x_1) = x(t)

Is this the system of linear equations I am looking for?

 

[Chestermiller]

Thank you for the reply. Are my FBDs right though?

So assuming that they are:
For node 1: k(x_1) = x(t)
For node 2: 6k(x_2 – x_1) = x(t)
For node 3: 30k/11(x_3 – x_1) = x(t)

Is this the system of linear equations I am looking for?

No. For the force balance on node 1, I get:
$$kx_1=5k(x_2-x_1)+k(x_3-x_1)$$
or, equivalently,
$$7x_1-5x_2-x_3=0$$
Can you do similar force balances for nodes 2 and 3?

Chet

 

[mintsnapple]

I'm having a hard trouble understanding. Could you help me please draw a FBD for node 1? I thought that the force went through the nodes, so that the force balance is just k(x_1) = x(t). Thank you so much.

 

[Chestermiller]

I'm having a hard trouble understanding. Could you help me please draw a FBD for node 1? I thought that the force went through the nodes, so that the force balance is just k(x_1) = x(t).

Actually, no. Choosing Node 1 as a free body, there are 4 springs connected to node 1. These are exerting the only contact forces on node 1. One of them has a spring constant of k, and is connected to the wall. One has a spring constant of k, and is connected to node 3. Two of them have spring constants of 2k and 3k, respectively, and are connected to node 2. Do you see which springs I'm referring to? In terms of the nodal displacements x1, x2, and x3, what are the tensions in each of these springs? Write a force balance on node 1, based on these tensions.

Chet

 

[mintsnapple]

Ah. This makes everything so much more clear. So then the forces of the three springs to the right of node 1 are opposite in direction to the force of the spring to the left.

The parallel springs connected to node 2 have a k_eq of 2k = 3k = 5k. The tension in this spring is equal to then, 5k(x_2 - x_1).

The spring connected to node 3 has a tension equal to k(x_3-x_1).
Thus, the force balance on node 1 is k(x_1) = 5k(x_2 - x_1) + k(x_3-x_1).
--
Now, for node 2, two springs to the left are connected to node 1, and 1 spring on its right is connected to node 3.
Thus the force balance is 5k(x_1 - x_2) = 5k(x_3 - x_2).

--
For node 3, one spring is connected to node 2, one is connected to node 1, and one is connected to x(t).
Thus, the force balance is 5k(x_2 - x_3) + k(x_1 - x_3) = k(x_3) + x(t).

Does this seem correct?

 

[Chestermiller]

It looks like you have the right idea now (nice work), but there are some sign errors.

Node 2: 5k(x_2 - x_1) = 5k(x_3 - x_2)
Node 3: 5k(x_3 - x_2) + k(x_3 - x_1) = k(x(t)-x_3)

Chet

 

[mintsnapple]

Thank you so much! But one more thing: I was taught that the force due to a spring is given by k(x_near - x_far), where x_near is the node whose restoring force is being considered. Could you clarify that concept for me a bit more in terms of this problem? I would very much appreciate it. Take for example Node 1. The tension in the spring connected to node 3 is given by k(x_3 - x_1). Why would this not be k(x_1 - x_3)? Are we not taking node 1 into consideration?

 

[Chestermiller]

Thank you so much! But one more thing: I was taught that the force due to a spring is given by k(x_near - x_far), where x_near is the node whose restoring force is being considered. Could you clarify that concept for me a bit more in terms of this problem? I would very much appreciate it. Take for example Node 1. The tension in the spring connected to node 3 is given by k(x_3 - x_1). Why would this not be k(x_1 - x_3)? Are we not taking node 1 into consideration?

All three of the displacements x1, x2, and x3 are taken as positive to the right. What we need to do is calculate the increase in length of each of the springs based on the displacements at its two ends. This determines the tension in the spring. So, if a spring is connected between nodes 1 and 3, for example, its increase in length is x3 - x1. In this problem, it's the displacement of the node at the right minus the displacement of the node at the left.

Chet

 

[mintsnapple]

So if I say k(x_1) = 5k(x_2 - x_1) + k(x_3-x_1).
Is it right to say that x_1 = (5x_2 + x_3) / 7? Or should I express it some other way?

 

[Chestermiller]

Good morning. I'm back.

So if I say k(x_1) = 5k(x_2 - x_1) + k(x_3-x_1).
Is it right to say that x_1 = (5x_2 + x_3) / 7? Or should I express it some other way?

No. That's not what they are looking for. They are looking for x_1 expressed solely in terms of x(t).

You have 3 simultaneous linear algebraic equations in three unknowns: x_1, x_2, and x_3. Solve this set of equations for x_1, x_2, and x_3, each solely in terms of x(t). I'm certain that you have learned how to do a math problem like this.

Chet