Logarithmic Decrement Damping Simulator

Q: What is the logarithmic decrement?

The logarithmic decrement δ is the natural logarithm of the ratio of successive peak amplitudes in a free-vibration decay. Using the first peak amplitude A1 and the peak An that occurs n full cycles later, it is δ = (1/n)·ln(A1/An). Averaging over n cycles reduces the reading error compared with using a single cycle. The larger the damping, the faster the amplitude shrinks and the larger δ becomes. This tool computes δ from two peak amplitudes and the number of elapsed cycles.

Q: How do I get the damping ratio from the logarithmic decrement?

The damping ratio ζ follows exactly from the log decrement δ as ζ = δ / √(4π² + δ²). For light damping (roughly ζ < 0.1) the simple approximation ζ ≈ δ/(2π) is accurate enough. For example, δ = 0.285 gives ζ ≈ 0.285/6.283 ≈ 0.0454, almost identical to the exact value. As damping grows, the δ² term in the denominator can no longer be neglected, so this tool always uses the exact formula.

Q: How does the damped natural frequency differ from the undamped one?

The undamped natural frequency fn is the frequency the system would have with no damping at all, while the damped natural frequency fd is the frequency at which the damped system actually oscillates. They are related by fd = fn·√(1−ζ²), so damping makes the oscillation slightly slower. For light damping (ζ < 0.1) the gap between fd and fn is less than 1% and is usually negligible, but it becomes noticeable once ζ exceeds about 0.3. The damped period Td = 1/fd can be measured directly as the time between adjacent peaks of the decay.

Q: How is the quality factor Q related to the damping ratio?

The quality factor is a dimensionless measure of how sharp a system is, related to the damping ratio by Q = 1/(2ζ). The smaller the damping, the larger Q and the sharper the resonance. For example, ζ = 0.045 gives Q ≈ 11 and ζ = 0.01 gives Q = 50. Q can also be read as the ratio of energy stored to energy lost per cycle, and it can be obtained from a frequency response as Q = fn/Δf using the half-power bandwidth. Checking that Q from the log decrement matches Q from the frequency response is a good sanity check on the measurement.

Parameters

First peak amplitude A₁

Height of the first peak in the decay waveform

Peak amplitude after n cycles Aₙ

Peak height read n cycles after the first one

Elapsed cycles n

cycles

Number of full cycles between A₁ and Aₙ

Undamped natural frequency fₙ

Frequency assuming no damping at all

Results

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Log decrement δ

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Damping ratio ζ

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Damped nat. freq. fd (Hz)

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Amplitude ratio / cycle

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Quality factor Q

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Damping class

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Free-vibration decay — drawing animation

The decaying sinusoid x(t) is drawn with its exponential envelope ±A·e^(−ζωₙt) (dashed). A marker tracks the trace and the first peak A₁ and the peak Aₙ after n cycles are highlighted.

Free-vibration decay waveform x(t)

Peak amplitude decay (logarithmic y-axis)

Theory & Key Formulas

$$\delta=\frac{1}{n}\ln\frac{A_1}{A_{n}},\qquad \zeta=\frac{\delta}{\sqrt{4\pi^2+\delta^2}}$$

Logarithmic decrement δ and damping ratio ζ. A₁: first peak amplitude, Aₙ: peak amplitude after n cycles, n: elapsed cycles.

$$f_d=f_n\sqrt{1-\zeta^2},\qquad Q=\frac{1}{2\zeta}$$

Damped natural frequency fd (the actual oscillation frequency) and quality factor Q. fₙ: undamped natural frequency.

For light damping (ζ < 0.1) the approximation ζ ≈ δ/(2π) holds, so the damping ratio can be estimated from δ in your head.

What is logarithmic decrement damping measurement?

🙋

The damping ratio tells you how quickly a vibration dies out, right? But how do you actually measure it? The damping coefficient c in the equations seems impossible to read off a real part.

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Good question. Measuring the damping coefficient c directly really is hard. That is why we use the "logarithmic decrement" idea instead. The procedure is simple: tap the object once and let it vibrate freely, then just record the waveform as the motion dies away. How much each peak shrinks from one cycle to the next tells you the damping.

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The "shrinkage" of the peak heights — so I just compare two neighbouring peaks?

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Exactly. The envelope of a damped free vibration decays as the exponential e^(−ζωₙt), so each peak shrinks by the same ratio every cycle. The natural logarithm of the ratio of adjacent peaks A₁/A₂ is the logarithmic decrement δ. But adjacent peaks differ so little that reading error creeps in. So in practice we compare peaks n cycles apart: δ = (1/n)·ln(A₁/Aₙ). Raise the "elapsed cycles n" on the left and the measurement becomes more stable because it is averaged.

🙋

I see — dividing by n brings it back to a per-cycle figure. And how do you get the damping ratio ζ from δ?

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Exactly, they are linked by ζ = δ/√(4π²+δ²). In the field people usually just remember the rougher ζ ≈ δ/(2π) — for light damping the two are practically the same. With the default values, δ ≈ 0.285 and ζ ≈ 0.045. Steel structures and well-built machines usually land around ζ = 0.01 to 0.05. If ζ exceeds 0.2, you are looking at a system where rubber or a dedicated damper is doing real work.

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There's also a "quality factor Q" in the results. What does that represent?

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Q measures the "sharpness of the resonance", through the simple relation Q = 1/(2ζ). The smaller the damping, the larger Q and the more needle-sharp the resonance peak. ζ = 0.045 gives Q ≈ 11. Things that "ring well", like a loudspeaker or a tuning fork, have a high Q; things you want to stop quickly, like a seismic damper, are deliberately given a low Q. Think of damping ratio, logarithmic decrement and Q as three different words for the same damping property.

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One last thing. Is it true the frequency changes when there is damping? I assumed the natural frequency was fixed.

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It is true. Damping makes the oscillation slightly slower. Against the undamped natural frequency fₙ (assuming no damping), the system actually oscillates at the damped natural frequency fd = fₙ·√(1−ζ²). For small ζ, √(1−ζ²) is almost 1 and the gap is under 1% — even with the default values fd ≈ 9.99 Hz, barely different from fₙ = 10 Hz. So most practical calculations just take fd ≈ fₙ. But once ζ grows to 0.3 or 0.5 the gap can no longer be ignored, so keep it in the back of your mind.

Frequently Asked Questions

The logarithmic decrement δ is the natural logarithm of the ratio of successive peak amplitudes in a free-vibration decay. Using the first peak amplitude A1 and the peak An that occurs n full cycles later, it is δ = (1/n)·ln(A1/An). Averaging over n cycles reduces the reading error compared with using a single cycle. The larger the damping, the faster the amplitude shrinks and the larger δ becomes. This tool computes δ from two peak amplitudes and the number of elapsed cycles.

The damping ratio ζ follows exactly from the log decrement δ as ζ = δ / √(4π² + δ²). For light damping (roughly ζ < 0.1) the simple approximation ζ ≈ δ/(2π) is accurate enough. For example, δ = 0.285 gives ζ ≈ 0.285/6.283 ≈ 0.0454, almost identical to the exact value. As damping grows, the δ² term in the denominator can no longer be neglected, so this tool always uses the exact formula.

The undamped natural frequency fn is the frequency the system would have with no damping at all, while the damped natural frequency fd is the frequency at which the damped system actually oscillates. They are related by fd = fn·√(1−ζ²), so damping makes the oscillation slightly slower. For light damping (ζ < 0.1) the gap between fd and fn is less than 1% and is usually negligible, but it becomes noticeable once ζ exceeds about 0.3. The damped period Td = 1/fd can be measured directly as the time between adjacent peaks of the decay.

The quality factor is a dimensionless measure of how sharp a system is, related to the damping ratio by Q = 1/(2ζ). The smaller the damping, the larger Q and the sharper the resonance. For example, ζ = 0.045 gives Q ≈ 11 and ζ = 0.01 gives Q = 50. Q can also be read as the ratio of energy stored to energy lost per cycle, and it can be obtained from a frequency response as Q = fn/Δf using the half-power bandwidth. Checking that Q from the log decrement matches Q from the frequency response is a good sanity check on the measurement.

Real-World Applications

Modal testing of buildings and civil structures: Bridges, high-rise buildings and chimneys are made to vibrate freely by wind, traffic or a shaker, and the logarithmic decrement is extracted from the decaying waveform. The damping ratio of steel structures is typically ζ = 0.005 to 0.02, and 0.02 to 0.05 for reinforced concrete. In seismic and wind design, this measured damping ratio is used directly as an input to the response calculation, so getting less damping than the design value can lead to unexpectedly large motions.

Damping evaluation of machine parts and rotating machinery: For machine-tool spindles, turbine blades and piping systems, the impulse test — striking the part once with an impact hammer and recording the response — is widely used. The logarithmic decrement is obtained from the peak decay, and the resonant amplification factor (Q) is estimated, which shows how dangerous operation near a critical speed would be. A part with a high Q (low damping) sees its amplitude shoot up the instant it enters resonance.

Verifying vibration-control and isolation designs: If you measure the logarithmic decrement before and after adding rubber mounts, oil dampers or damping alloys, you can quantitatively show how much the countermeasure raised the damping ratio. The strength of the log-decrement method is being able to say "δ rose from 0.05 to 0.25" with numbers rather than "the vibration seems lower". The same technique is used when tuning car suspensions and engine mounts.

Validating CAE damping models: Vibration analysis by the finite element method needs a damping ratio entered as Rayleigh damping or modal damping. Guessing this input badly throws the response calculation far off. Impulse-testing the real machine, extracting ζ from the logarithmic decrement and feeding that value into the analysis model — this "model calibration" is essential for trustworthy vibration simulation.

Common Misconceptions and Pitfalls

The most common problem is being vague about where the peak amplitudes are read. The logarithmic decrement must compare peaks of the same sign and the same direction. If you mix positive crests with negative troughs, or mistake one cycle for a half-cycle in a heavily damped system, δ comes out twice or half the true value. Furthermore, the first few cycles right after the strike often still carry the transient disturbance of the impulse and contamination from higher modes, so using that region as A₁ adds error. The trick is to pick a stable, clean cycle as A₁ and a peak Aₙ that has decayed substantially but is not yet buried in measurement noise.

Next, it is easy to forget that the logarithmic decrement assumes a single-degree-of-freedom system. This method only holds when the decay waveform is a clean, single-frequency exponentially decaying sinusoid. In a multi-degree-of-freedom system where several closely-spaced natural frequencies are excited together, the waveform shows "beats" and the peak heights do not decrease monotonically, so applying the logarithmic decrement naively gives a wrong value. On real machinery the proper approach is to extract only the target mode with a band-pass filter before computing δ, or to estimate the exponential envelope by curve fitting. Note that this tool deals with an ideal single-degree-of-freedom, viscously damped system.

Finally, there is the assumption that damping is constant regardless of amplitude. With the viscous damping this tool assumes, the damping ratio is constant regardless of amplitude and the logarithmic decrement is the same value every cycle. But real damping often contains "Coulomb friction damping (constant damping)" or "hysteretic damping" from bolted-joint slip or internal material friction, and these change how damping acts as amplitude varies. When friction damping dominates, the peaks decay linearly rather than exponentially, and the apparent logarithmic decrement grows larger as the amplitude shrinks. If the decay waveform does not fall on a straight line in a semi-log plot, that is a sign you should question the viscous damping assumption itself.

What is logarithmic decrement damping measurement?

Frequently Asked Questions

Real-World Applications

Common Misconceptions and Pitfalls

How to Use

Worked Example

Practical Notes