# Damping: Three Basic Questions

1. Nov 5, 2012

### mudbone

Hello All,

I have some basic damping related questions that I am hoping you all can shed some light on or suggest some resources on to someone who has taken a few undergraduate level physics courses.

i.) Can someone give me an intuitive explanation as to why linear damping (underdamped case) of a harmonic oscillator shortens the period?

ii.) Must an oscillatory system with any kind of damping (including nonlinear damping) have a shorter period/higher frequency than the undamped system?

iii.) Why does damping cause a broader peak in the power spectrum of the oscillatory time series (please try to give me an intuitive reason not a purely mathematical one)?

Thank you in advance

-Mudbone

2. Nov 5, 2012

### Philip Wood

Increasing the damping lengthens the period (lowers the frequency). The effect is very small unless the damping is very heavy; for example even if the damping halves the oscillation amplitude every cycle, the period is lengthened by less than 1 per cent.

Why does damping lengthen the period? Well, a hand-waving explanation would be along the lines that damping impedes the motion...

Your third question is perhaps the easiest to answer. Far from resonance the amplitude is low, so the speed of motion is low, so the damping forces are smaller. Thus extra damping has less effect on the amplitude far from resonance than it does nearer resonance, when the body is (on average) moving faster. So the resonance curve is depressed more in the middle than at the edges, that is, it is broadened.

3. Nov 5, 2012

### AlephZero

Sightly less hand-waving: Suppose you have an undamped oscillator that exchanges potential energy and kinetic energy in each cycle of vibration.

If you add some damping (of any type), you take energy out of the system. So if you start with the same amount of PE as the undamped system, by the time the PE has reduced to 0 the amount of KE in the system is smaller. So the velocity is smalller. So the system takes longer to get to the position where the PE = 0. So the frequency is lower.

4. Nov 5, 2012

### nasu

A sharp, narrow peak in the frequency spectrum corresponds to a single sine or (cosine) wave. Still has some width due to the fact that the signal is finite in time.
The damped oscillation is not a pure sine but is exponentially decreasing. How can you get that? One way would be to add many sine waves with various frequencies and amplitudes. These are represented by the wide peak in the frequency spectrum.

5. Nov 6, 2012

### Philip Wood

AlephZero. Got it! You're considering a quarter cycle at a time. Nice! [I was going to object that although the mean speed is less, so too is the amplitude, but if we consider each quarter of a cycle at a time, the objection disappears.]

6. Nov 6, 2012

### K^2

Lets look at the "quarter"-cycle from the node in displacement until KE goes to zero. Now consider that "quarter"-cycle backwards in time. Your logic applies, except now energy is added to the system. So while any kind of damping will increase the time it takes for system to return from maximum displacement, it reduces the amount of time it takes to go from zero displacement to the turning point. (Which makes sense. Same starting KE, less distance to travel.)

For linearly damped oscillator, the net effect is increased period. I am not convinced, on this argument alone, that this is true for a general case. That said, I cannot come up with an example where period is not increased, so there is probably a way to prove this for general case.

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