Mass–Spring–Damper (δ and ω form) Summary

0.1 Notes

Regarding the second order–homogenous ODE

my + by + ky = 0 =⇒ y + 2δy + ω2y = 0

where km

= ω2 (ω denotes angular frequency) and bm

= 2δ (δ denotes damping) thenthe general solution for the position y(t) of the spring is

y(t) =

C1e(−δ+

√δ2−ω2)t + C2e

(−δ−√δ2−ω2)t when δ2 > ω2

e−δt(C1 cos(√ω2 − δ2t) + C2 sin(

√ω2 − δ2t) when δ2 < ω2

C1e−δt + C2te

−δt when δ2 = ω2.

The velocity v(t) of the spring is found by computing y(t), i.e. by differentiating y(t).The constants C1 and C2 are found by solving the system of equations y(0) = y0and v(0) = v0 where x0 and v0 are the given initial position and initial velocity,respectively, of the spring. You should be able to understand the form of the solutionsy(t) in each of the 3 cases based on the eigenvalues of the matrix

A =

(0 1−ω2 −2δ

).

The cases δ2 > ω2, δ2 < ω2, δ2 = ω2 correspond to the overdamped, underdamped,and critically damped motion, respectively, of the spring.

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