# What's the opposite of damping?

When a function (for instance, a sine wave) is multiplied by a decaying exponential, we call the phenomenon damping. What would it be called if the wave is multiplied by a growing exponential?

I suggested forcing or driving, since functions of that form arise in simple harmonic oscillators driven by sine waves with frequency equal to the oscillator's frequency. But I wondered if others had words for it.

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"Negative damping" or "positive feedback" are common. –  Arturo Magidin Apr 27 '11 at 17:13
Driven/driving is certainly the term I've heard; see here for example. –  Zev Chonoles Apr 27 '11 at 17:15
Exciting comes to mind. –  Emre Apr 27 '11 at 17:20
Ramping? ...... –  mjqxxxx Apr 27 '11 at 19:24
People (at least in electrical engineering) use amplification (=gain) as the opposite of damping (=loss)... –  Fabian Apr 27 '11 at 20:13

I would call reinforcing and amplifying a signal positive feedback, and where you have neither dampening nor feedback I would call that resonance or ringing.

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In physics, it is indeed called forcing or driving. For example, the solutions to a spring problem satisfy the spring equation (the solutions are sines). If you apply an external force to the object attached to the spring, depending on the nature of the force, it can have a driving effect on the object. If you for example apply a sinusoidal, periodic force with the same frequency as the characteristic frequency of the spring system, you shall see that the solution will grow in magnitude for increasing $t$.

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I disagree. In the typical spring-mass-damper, $m\ddot{x} + c\dot{x}+kx = F(x,t)$, the "forcing" term is $F(x,t)$, but the damping coefficient term is $c$. Changing signs on $c$ such that the damping effect is reversed is not the same as adding a forcing function. –  Arkamis Aug 11 '14 at 16:05
You are right, I was trying to explain it in physical terms and in mechanical systems negative damping does not exist so you have to resort to some sort of external driving force. If you managed to apply a driving force $F(x,t) = k \dot{x}$ $k > c$ then the effective damping would be negative but I am not sure if you can realize that. –  hickslebummbumm Aug 11 '14 at 16:24

isn't "amplification" more appropriate?

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