Spring-mass system with nearest neighbor couplings

[sapiental]

Hello,

This problem refers particulary to the model in my attachment.

The blocks are free to move but only along the x-axis. The springs connecting adjacent blocks have spring constant k1, while the two outer springs have stiffness k2. All the springs have a rest lengt of one. Write the matrix equation for the equilibrium position of the blocks and plot the total length of the system as a function of k1/k2

The model of the problem is throwing me off here because I am used to spring systems being fixed to a wall at one end and having something else attached to it on the other end. The way I'm approaching the problem is by finding out the individual force on each block.. F1 + F2 + F3 + F4. However, I keep getting stuck in a loop by trying to focus on an individual block because each block seems dependent on the other ones at all times.

I also can't picture the equation f(k1/k2) at all in my head.

I know it's not much but this is what I have so far.

Springs in a parallel
kp = k1+k2
Springs in a series
1/ks = 1/k1 + 1/k2

The length of the system should be related to the equilibrium length when all springs are at rest.

so:

L_equilibrium = L_1e + L_2e + L_3e

(these are the equilibrium distances between each block)

I'm guessing that the force on each end is something like:

F = -k_springtotal(L_system - L_equilibrium)I would really appreciate it if someone could give me a push start with this problem. I mainly want to understand what force the individual blocks are experiencing if displaced. Also, I don't really understand the question that well either to be completely honest. Particulary the part about finding the length based on the ratio of constants? How do I get the individual forces on each block if I'm not given specific constants for k1 and k2?

Any help is greatly appreciated.

 

[AlephZero]

Number the masses 1 2 3 4 from left to right and call the positions x1 x2 x3 x4

For the top spring, the length is x4-x2, the extension is x4-x2-1, and the forces are -k2(x4-x2-1) applied to block 4 and +k2(x4-x2-1) applied to block 2.

The total force on eack block is the just the sum of the forces from all the springs connected to it. That will be an expression of the form a1 x1 + a2 x2 + a3 x3 + a4 x4 + c, and it equals 0 since the system is in equilibrium.

You are not asked to find the individual forces, only the displacements. Obviously the forces depend on the values of k1 and k2 but the displacements will only depend on the ratio k1/k2.

 

[m2003]

Thank you for sharing your thoughts and questions on this spring-mass system with nearest neighbor couplings.

To start, let's define the variables:

- m = mass of each block
- k1 = spring constant for the springs connecting adjacent blocks
- k2 = spring constant for the outer springs
- x1, x2, x3, x4 = displacements of each block from their equilibrium positions

Now, let's write the equations of motion for each block:

For block 1:
m(d^2x1/dt^2) = -k1(x1-x2) - k2x1

For blocks 2 and 3:
m(d^2x2/dt^2) = -k1(x2-x1) - k1(x2-x3)
m(d^2x3/dt^2) = -k1(x3-x2) - k1(x3-x4)

For block 4:
m(d^2x4/dt^2) = -k1(x4-x3) - k2x4

Next, let's rearrange the equations to get them in matrix form:

| m 0 0 0 | | d^2x1/dt^2 | = | -k1 -k2 0 0 | | x1 |
| 0 m 0 0 | | d^2x2/dt^2 | = | -k1 2k1 -k1 0 | | x2 |
| 0 0 m 0 | | d^2x3/dt^2 | = | 0 -k1 2k1 -k1 | | x3 |
| 0 0 0 m | | d^2x4/dt^2 | = | 0 0 -k1 -k2 | | x4 |

We can then simplify this to:

M * d^2x/dt^2 = -K * x

where M is the mass matrix and K is the stiffness matrix.

To find the equilibrium positions, we can set the second derivative terms to zero and solve for x:

M * 0 = -K * x
x = -M^-1 * K * x

This gives us the equilibrium positions for each block in terms of the stiffness constants k1 and k2. We can then plot the total length of the system (L