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Step response of a damped harmonic oscillator; curves are plotted for three values of μ = ω 1 = ω 0 √ 1 − ζ 2. Time is in units of the decay time τ = 1/(ζω 0). The value of the damping ratio ζ critically determines the behavior of the system. A damped harmonic oscillator can be:
The equation describes the motion of a damped oscillator with a more complex potential than in simple harmonic motion (which corresponds to the case = =); in physical terms, it models, for example, an elastic pendulum whose spring's stiffness does not exactly obey Hooke's law.
Two such solutions, for the two values of s satisfying the equation, can be combined to make the general real solutions, with oscillatory and decaying properties in several regimes: Phase portrait of damped oscillator, with increasing damping strength. It starts at undamped, proceeds to underdamped, then critically damped, then overdamped. Undamped
This plot corresponds to solutions of the complete Langevin equation for a lightly damped harmonic oscillator, obtained using the Euler–Maruyama method. The left panel shows the time evolution of the phase portrait at different temperatures. The right panel captures the corresponding equilibrium probability distributions.
In the classical approach, the Rabi problem can be represented by the solution to the driven damped harmonic oscillator with the electric part of the Lorentz force as the driving term: ¨ + ˙ + = (,),
A highly damped circuit will fail to resonate at all, when not driven. A circuit with a value of resistor that causes it to be just on the edge of ringing is called critically damped. Either side of critically damped are described as underdamped (ringing happens) and overdamped (ringing is suppressed).
If a system initially rests at its equilibrium position, from where it is acted upon by a unit-impulse at the instance t=0, i.e., p(t) in the equation above is a Dirac delta function δ(t), () = | = =, then by solving the differential equation one can get a fundamental solution (known as a unit-impulse response function)
In other words, the solution of equation 2, u(x), can be determined by the integration given in equation 3. Although f ( x ) is known, this integration cannot be performed unless G is also known. The problem now lies in finding the Green's function G that satisfies equation 1 .