Part 1 · The elementary oscillator

1.4 Measuring damping

Source: Carboni’s notes for lesson 1, Oscillatore elementare, pp. 8–10

Objectives

Recipe. The damping ratio from two successive peaks of a free oscillation (logarithmic decrement).
Recipe. The damping ratio from the areas of one force–displacement cycle.
Picture. $ξ$ as the energy lost in a cycle compared with the energy stored.

$K$ and $M$ can be weighed and measured; the damping $C$ usually cannot, because it comes from many small losses (air, internal friction, joints). It is measured on the system itself, from its response. The lesson gives two ways.

The logarithmic decrement

Record a free oscillation of an underdamped system (Key observation 1.3.6) and compare two successive peaks. The response at two instants one damped period apart is $U (t_{n}) = A e^{- ξ ω t_{n}} cos (ω_{d} t_{n} - ϕ), U (t_{n + 1}) = A e^{- ξ ω (t_{n} + T_{d})} cos [ω_{d} (t_{n} + T_{d}) - ϕ] .$ The two cosines are equal, because $ω_{d} T_{d} = 2 π$ adds a whole turn to the angle (beyond the lectures: this is why they cancel). Only the envelope differs.

Definition 1.4.1decrementnotes L1, p. 8

The decrement is the ratio of two successive peaks: $δ = \frac{U ( t _{n} )}{U ( t _{n + 1} )} .$

Proposition 1.4.2damping from the decrementnotes L1, pp. 8–9

$ln δ = \frac{2 π ξ}{1 - ξ ^{2}}, so ξ = \frac{ln δ}{4 π ^{2} + ( ln δ ) ^{2}} and, for ξ ≪ 1, ξ ≃ \frac{ln δ}{2 π} .$

Proof

The ratio, with the cosines cancelled: $δ = \frac{A e ^{- ξ ω t_{n}}}{A e ^{- ξ ω (t_{n} + T_{d})}} = e^{ξ ω T_{d}} .$
Substitute $T_{d} = \frac{2 π}{ω _{d}} = \frac{2 π}{ω 1 - ξ ^{2}}$ ; the $ω$ cancels: $δ = e^{\frac{2 π ξ}{1 - ξ ^{2}}} .$
Take the logarithm: $ln δ = \frac{2 π ξ}{1 - ξ ^{2}}$ .
Light damping. If $ξ ≪ 1$ , $1 - ξ^{2} ≃ 1$ and $ξ ≃ ln δ /2 π$ .
Exactly. Square and multiply by $1 - ξ^{2}$ : $(ln δ)^{2} (1 - ξ^{2}) - 4 π^{2} ξ^{2} = 0 ⟹ ξ^{2} [(ln δ)^{2} + 4 π^{2}] = (ln δ)^{2},$ and take the positive root.

This is the logarithmic decrement method (metodo del decadimento logaritmico).

Loading the demo…

Figure 1.4.1 A recorded free decay with its peaks marked. The readout takes the first two peaks, computes

δ

and

ln δ

, and gives

ξ

back with both formulas. Raise

ξ

to see the approximation drift away from the true value.

Warningslip in the notes

In step 2 the notes write $1 - ξ$ in the exponent; the square on $ξ$ is missing, and the next line has $1 - ξ^{2}$ again. The notes also write the damped period both $T_{d}$ and $T_{D}$ ; they are the same.

Remarkbeyond the lectures

In practice, use many periods. Peaks $N$ periods apart have the ratio $U (t_{n}) / U (t_{n + N}) = δ^{N}$ , so $ln δ = \frac{1}{N} ln \frac{U ( t _{n} )}{U ( t _{n + N} )} .$ Averaging over many periods reduces the effect of noise on a single pair of peaks.

From the force–displacement cycle

The second way uses one measured cycle, like the right-hand plot of Figure 1.3.1. Its two areas give the two constants (Proposition 1.3.2): $W_{E} = \frac{1}{2} K A^{2} ⟹ K = \frac{2 W _{E}}{A ^{2}}, W_{D} = π C Ω A^{2} ⟹ C = \frac{W _{D}}{π Ω A ^{2}} .$

Proposition 1.4.3damping from a cyclenotes L1, p. 10

$ξ = \frac{W _{D}}{4 π W _{E}} \frac{ω}{Ω}, and at resonance (Ω = ω) : ξ = \frac{W _{D}}{4 π W _{E}} .$

Proof

Write $C_{cr}$ with $K$ only, multiplying by $ω / K / M = 1$ : $ξ = \frac{C}{C _{cr}} = \frac{C}{2 K M} \cdot \frac{ω}{K / M} = \frac{C ω}{2 K} .$
Substitute $C$ and $K$ from the areas: $ξ = \frac{W _{D}}{π Ω A ^{2}} \cdot \frac{A ^{2}}{4 W _{E}} ω = \frac{W _{D}}{4 π W _{E}} \frac{ω}{Ω} .$
Run the cycle at resonance, $Ω = ω$ , and the frequency ratio is 1.

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Figure 1.4.2 The cycle method. With

M = 1

ω = K

; the readout computes

ξ

from the two areas and checks it against

C /2 ω M

. Set

Ω

ω

for the resonance formula.

Warningslip in the notes

p. 10 writes $C_{cr} = 2 K m$ and $K / m$ with a small $m$ ; it is the oscillator’s mass $M$ (the small $m$ was the added mass of the MEMS example). The middle step also writes $\frac{W _{D}}{π A ^{2}}$ , dropping the $Ω$ that the last fraction puts back.

Key observation 1.4.4what the damping ratio measuresbeyond the lectures

$ξ$ compares the energy lost in one cycle with the energy stored: at resonance $W_{D} / W_{E} = 4 π ξ$ . A lightly damped system with $ξ = 0.01$ loses about 13% of its stored energy per cycle.

Exercises

Exercise 1.4.1

In a free-decay test two successive peaks are 10 mm and 7 mm. Find $ξ$ with the approximate and the exact formula.

Solution

$δ = 10/7$ , $ln δ = 0.3567$ .
Approximate: $ξ ≃ 0.3567/2 π = 0.05677$ .
Exact: $ξ = 0.3567/ 4 π^{2} + 0.356 7^{2} = 0.05668$ . They differ by 0.2%: for a $ξ$ of a few percent the approximation is enough.

Exercise 1.4.2

A cycle test gives $W_{E} = 2$ J and $W_{D} = 1.5$ J. Find $ξ$ if the test ran at resonance, and if it ran at $Ω = ω /10$ .

Solution

At resonance: $ξ = 1.5/ (8 π) = 0.0597$ .
At $Ω = ω /10$ : $ξ = 0.0597 \cdot 10 = 0.597$ . The same dissipated energy at a tenth of the frequency means a ten times larger $C$ , since $W_{D} = π C Ω A^{2}$ .

Exercise 1.4.3

For the oscillator with $M = 2$ kg, $K = 800$ N/m, $C = 8$ N·s/m ( $ξ = 0.1$ ), find the decrement $δ$ . After how many periods has the amplitude halved?

Solution

$δ = e^{2 π ξ / 1 - ξ^{2}} = e^{2 π \cdot 0.1/ 0.99} = e^{0.6315} = 1.880$ .
Halving: $δ^{N} = 2$ , so $N = ln 2/ ln δ = 1.10$ periods.

Exercise 1.4.4

Peaks of a free decay are read at $t_{0}$ and ten periods later: 12.0 mm and 4.5 mm. Estimate $ξ$ .

Solution

$ln δ = \frac{1}{10} ln (12/4.5) = \frac{1}{10} \cdot 0.9808 = 0.09808$ .
$ξ ≃ 0.09808/2 π = 0.0156$ ; the exact formula gives the same to three digits.