Coupled Oscillator: Pulley

From Class Wiki
Jump to navigation Jump to search

Coupled Oscillator: Pulley System

Problem

Given the system shown, with all initial positions set so that gravity is accounted for and no motion is occurring, find the response. The spring around the pulley is modeled as a single spring with constant k2.

Error creating thumbnail: File missing


















Solution

We begin by writing the sum of forces equations for both blocks.

On the left, we have

F=ma

k1x1k2(x1x2)=m1x1¨

x1(k1+k2)x2k2m1=x1¨

On the right, we have

F=ma

k2(x2x1)=m2x2¨

k2(x2x1)m2=x2¨

So, our state space matrices are

[x1¨x1˙x2¨x2˙]=[0k1+k2m10k2m110000k2m20k2m20010][x1˙x1x2˙x2]

Now, suppose we impose the initial conditions as follows:
m1=10 kg
m2=20 kg
k1=10 N/m
k2=50 N/m

We now have

[x1¨x1˙x2¨x2˙]=[0605100002.502.50010][x1˙x1x2˙x2]

Using any number of different computer utilities, we can calculate the eigenvalues and eigenvectors of the coefficient matrix quickly and easily.

λ1=0.705+j2.179
λ2=0.705j2.179
λ3=0.705+j2.179
λ4=0.705j2.179

k1[0.7480.101j0.3110.262j0.4600.226+j0.047]
k2[0.7480.101+j0.3110.262+j0.4600.226j0.047]
k3[0.7480.101j0.3110.262+j0.4600.226+j0.047]
k4[0.7480.101+j0.3110.262j0.4600.226j0.047]

The full solution is

x¯=c1k1eλ1t+c2k2eλ2t+c3k3eλ3t+c4k4eλ4t

Eigenmodes

The four distinct eigenvalues mean there will be four eigenmodes (though two are similar). They are

1. The two masses moving in the same direction at all times (i.e. one up and the other down). 2. The two masses moving in opposite directions at all times (i.e. both up or down). 3. and 4. The masses moving at plus or minus 90 degrees of phase offset.