# Total energy of a damped oscillator

#### Signifier

Is it possible to express the total energy of a damped linear oscillator as a function of time? I'm confused here. I'd like to find E(t). As the oscillation is damped, dE/dt should everywhere be negative (energy being dissipated as radiation or heat). By setting E(t) equal to zero, shouldn't I be able to solve for the time at which the energy of the oscillating system is zero, and thus the time at which the system stops oscillating? And shouldn't this time be finite?

Is there another way to find the time at which the damped oscillator will stop oscillating?

Thanks!

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#### marcusl

Gold Member
Yes. The peak amplitude of the oscillation, that is, the envelope, decays exponentially. Since average or rms energy is related to the peak amplitude, it also decays exponentially. In theory it never exactly reaches zero so you can't say when the oscillator "stops." In practice you can say it stops when the amplitude is comparable to thermal noise or some other criterion. It is more common to specify the time constant, which is the time for the envelope to decay to 1/e of its initial amplitude.

#### tim_lou

Well... you can simlpy solve the differential equation for a damped oscillator, then use
$$E=\frac{1}{2}kx^2+\frac{1}{2}m\dot{x}^2$$

#### Mindscrape

An ideal damped oscillator won't stop oscillating until infinite time has elapsed. However, practically the easiest way to find the time when the damped oscillator will top oscillating would be to determine the time constant, sqrt(m/k), and then multiply it by five because after five time constants the motion will be reduced to 1% (or something close to that) of its initial amplitude.

#### AlephZero

Homework Helper
As the oscillation is damped, dE/dt should everywhere be negative (energy being dissipated as radiation or heat).
The above posts are correct in saying the oscillation continues for ever, but dE/dt is not always negative.

Taking a mechanical damped oscillator for example, with equation of motion M x'' + C x' + K x = 0, the energy is dissipated by the the damper. The work is (force times velocity) = C x'^2. That is zero twice every cycle, when the velocity becomes zero.

If you evaluate E from tim_lou's equation you will get an exponential decay multiplied by a something looking like (A + B sin pt), which is a curve that "wobbles" around the "average" exponential decay in the energy.

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#### tim_lou

BTW, when you consider the different solutions for under-damped, over-damped or critically damped oscillation, you get different solution.

I think there won't be any wobbling in the over-damped or critically damped cases.