Damped, driven oscillations

This is the final post in a four-part series on vibrating systems. The first three parts were

and now we consider driven, damped vibrations. We are looking at the equation

m u″ + γ u′ + k u = F cos ωt

where all coefficients are positive.

As in the previous post, we need to find one solution to the equation with the forcing term F cos ωt and add it to the general solution to the homogeneous (free) equation.

The particular solution we’re looking for is simply R cos(ωt – φ), but the dependence of R and φ on the coefficients m, γ, and k is complicated. We’ll come back to this later, but first we’ll see what we can understand without knowing the exact form of R and φ.

The solutions to the unforced equation decay exponentially over time because of damping. The long-term behavior of the system, known as the steady state solution, is just R cos(ωt – φ). The frequency of the steady state solution is the same as the frequency of the forcing function, though the phase will be different in general.

To find the amplitude and phase of the steady state solution, we first define Δ > 0 by

Δ2 = m202 − ω2)2 + γ2 ω2

where ω0 = √(k/m) is the natural frequency of the system.

Then R = F /Δ . Note that Δ is small when the driving frequency ω is close to the natural frequency. That’s not where Δ is minimized, though when γ is small the minimum of Δ occurs close to the natural frequency. The minimum of Δ, and hence the maximum of the amplitude R, occurs at a lower frequency, namely

ω2 = ω02 – γ2 / 2m2.

The phase φ satisfies

cos φ = m02 − ω2)/ Δ


sin φ = γω/ Δ.

For low frequency ω, the phase φ is small, i.e. the response is nearly in phase with the excitation. For ω =ω0, i.e. forcing at the natural frequency, the response lags the excitation by π/2, a quarter cycle.

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One thought on “Damped, driven oscillations

  1. Nice.

    To be consistent (ok, picky) though, shouldn’t the post title be “Driven, Damped vibrations” rather than “Damped, driven oscillations”?

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