Transcript Physics 321 Hour 33 Coupled Oscillators I

Physics 321
Hour 33
Coupled Oscillators I
Example I
k1
x1
k2
x2
k3
1
1
2
𝑇 = 𝑚1 𝑥1 + 𝑚2 𝑥22
2
2
1
1
1
2
2
𝑈 = 𝑘1 𝑥1 + 𝑘2 𝑥2 − 𝑥1 + 𝑘3 𝑥22
2
2
2
1
1
1
2
2
𝐿 = 𝑚1 𝑥1 + 𝑚2 𝑥2 − 𝑘1 𝑥12
2
2
2
1
1
− 𝑘2 𝑥2 − 𝑥1 2 − 𝑘3 𝑥22
2
2
Example I
k1
x1
k2
x2
1
1
1
1
2
2
2
𝐿 = 𝑚1 𝑥1 + 𝑚2 𝑥2 − 𝑘1 𝑥1 − 𝑘2 𝑥2 − 𝑥1
2
2
2
2
k3
2
1
− 𝑘3 𝑥22
2
𝑚1 𝑥1 = −𝑘1 𝑥1 + 𝑘2 𝑥2 − 𝑥1 = − 𝑘1 + 𝑘2 𝑥1 + 𝑘2 𝑥2
𝑚2 𝑥2 = −𝑘3 𝑥2 − 𝑘2 𝑥2 − 𝑥1 = − 𝑘2 + 𝑘3 𝑥2 + 𝑘2 𝑥1
Example I
k1
𝑚1
0
x1
k2
𝑘1 + 𝑘2
0 𝑥1
=−
𝑚2 𝑥2
−𝑘2
𝐌𝑥 = −𝐊𝑥
x2
k3
𝑥1
−𝑘2
𝑘2 + 𝑘3 𝑥2
Normal Modes
Assume solutions of the form:
𝑥1 𝑡 = 𝐴1 sin 𝜔𝑡 + 𝜑1
𝑥2 𝑡 = 𝐴2 sin 𝜔𝑡 + 𝜑2
Note that 𝜔 is the same in both equations! – This
defines a “normal mode.”
Then 𝐌𝑥 = −𝜔2 𝐌𝑥 = −𝐊𝑥
Normal Modes - Method 1
𝐌𝑥 = −𝜔2 𝐌𝑥 = −𝐊𝑥
→ det 𝐊 − 𝜔2 𝐌 = 0
𝑘1 + 𝑘2 − 𝜔2 𝑚1
−𝑘2
=0
2
−𝑘2
𝑘2 + 𝑘3 − 𝜔 𝑚2
• Modified eigenvalue problem
• Solve for 𝜔2
𝑥1
2
• Use 𝐊 − 𝜔 𝐌 𝑥 = 0 to solve for 𝑥1 and 𝑥2
2
• Easiest by hand.
Normal Modes - Method 2
𝐌𝑥 = −𝜔2 𝐌𝑥 = −𝐊𝑥
𝐌−1 𝐊𝑥 = 𝜔2 𝑥
→ det 𝐌 −1 𝐊 − 𝜔2 𝟏 = 0
•
•
•
•
Standard eigenvalue problem
Eigenvalues are 𝜔2
Eigenvectors are normal modes
Easiest with Mathematica
Examples
coupled oscillators.nb
normal coords.nb