Undergraduate

UndergraduateClassical mechanicsOscillations and waves


Coupled oscillations and common modes


When two or more oscillating systems interact with each other, they can exhibit interesting and complex behavior. This interaction is often called coupling. A clear understanding of coupled oscillations is important because it underlies a variety of phenomena observed in classical and quantum physics as well as many technological applications.

Basic concepts

Let us consider two pendulums of the same length, connected by a spring. If you displace one pendulum and release it, you will see that after a while, the momentum is almost completely transferred to the other pendulum. This is a classic demonstration of coupled oscillations.

Fixed Support Spring

Mathematical analysis of coupled oscillations

Consider two coupled oscillators with equal masses m and spring constant k, connected by an additional spring with spring constant k_c. The equations of motion for displacements x1 and x2 from equilibrium can be written as:

m(d²x1/dt²) = -kx1 + kc(x2 - x1) m(d²x2/dt²) = -kx2 + kc(x1 - x2)

In matrix form, these equations become:

m(d²X/dt²) = -KX

where X is the vector of displacements and K is the matrix of spring constants:

X = | x1 | | x2 | K = | k + kc -kc | | -kc k + kc |

Normal mode

The normal mode of a coupled oscillatory system is a pattern of motion in which all parts of the system oscillate sinusoidally with the same frequency and maintain fixed relative amplitudes. Finding the normal mode of a system involves solving the eigenvalue problem:

(K - mω²I)A = 0

where ω is the angular frequency, I is the identity matrix, and A is the vector of amplitudes. The solutions to this problem give the allowed frequencies (eigenvalues) and the corresponding mode shapes (eigenvectors).

Example calculation

Suppose the mass of both pendulums is 1 kg, the spring constant is 50 N/m, and the coupling spring constant is 10 N/m:

K = | 60 -10 | | -10 60 |

The eigenvalue equation becomes:

| 60 - mω² -10 | | A1 | = 0 | -10 60 - mω² | | A2 |

Solving this for non-trivial solutions, we get:

(60 - mω²)² - 100 = 0

On solving the quadratic equation, we get two common mode frequencies.

Visualization of normal mode

For the calculated normal mode frequencies, we can look at their corresponding mode shapes:

1 2 Mode 1: In-phase 1 2 Mode 2: Out-of-phase

Energy in coupled oscillations

Energy may be transferred between the oscillators in a coupled system, producing beating or periodic energy oscillations between the modes. The total energy in such a system can be expressed as the sum of kinetic and potential energy. The individual energy expressions depend on the normal mode being excited.

E = T + U

where T is the kinetic energy and U is the potential energy. Analysis of energy transfer provides deep information about the dynamics of the system.

Application

Coupled oscillations and normal modes have wide applications in physics and engineering. Some notable examples include:

  • Molecular vibrations: Normal mode analysis helps understand the vibration spectrum of molecules, which is important for chemical reactions.
  • Mechanical systems: Engineering structures, including bridges and buildings, are designed with normal modes in mind to avoid resonant frequencies.
  • Electronics: Coupled circuits and devices such as filters and oscillators use the concepts of coupled oscillations for efficient design.

Conclusion

Understanding coupled oscillations and normal modes is fundamental to understanding complex oscillatory behavior in many physical systems. By studying these concepts, we can predict system behavior, optimize designs in engineering, and understand natural phenomena. The mathematical tools provided give general solutions applicable to many disciplines, demonstrating the versatility and importance of these classical mechanics principles.


Undergraduate → 1.8.3


U
username
0%
completed in Undergraduate


Comments