Acoustic properties · Resonance
Quality factor, decay and bandwidth
Does a longer ring always mean a larger Q?
Q describes energy retention per oscillation. Relating it to seconds of decay requires the mode’s frequency, while relating it to peak width requires a sufficiently isolated resonance.
Try the interactive demoStart with something familiar
Two swings can lose a similar fraction of motion on every pass, yet one makes more passes each second. Q counts retention per cycle, so “rings for more seconds” and “has higher Q” are not identical statements.
Amplitude in time
f = 500 Hz · Q = 50 / 150
Peak width
Green: Q = 50 · Blue: Q = 150
Notice what changes
Compare the same frequency first. At the same starting amplitude, larger Q means slower envelope decay. The curve shows amplitude, not the instantaneous up-and-down waveform.
Understand why
Now keep Q fixed and increase frequency. More cycles happen each second, so the same per-cycle retention results in a shorter decay measured in seconds.
Connect it to wood and making
In a frequency scan, the same weakly damped isolated mode appears as a peak. Higher Q means a narrower normalized peak. Nearby modes and recording windows can obscure this relationship in real tests.
Put names to the ideas
Energy per cycle
For weak damping, Q is approximately 2π times stored energy divided by energy lost per cycle. It is dimensionless, and is not a general quality certificate for wood.
Envelope and logarithm
An isolated weakly damped mode has amplitude envelope A₀ exp(−πft/Q). Natural-log amplitude is a line with slope −πf/Q; a power or energy log needs a different factor.
Half-power is not half-amplitude
Q ≈ f₀/Δf uses the full width between the two half-power points. Their amplitudes are 1/√2 of the peak, about −3 dB, not one half of peak amplitude.
τA is the time for amplitude to fall to about 37% of its start, not to zero. Δf is the full width of the resonance at 0.707 of peak amplitude; these are two ways to describe the same ideal mode’s damping.
A worked example
At 500 Hz and Q = 100, the amplitude time constant is about 64 ms and the approximate bandwidth is 5 Hz. Doubling frequency at the same Q halves the time constant.
Explore it yourself
Adjust frequency and Q separately. Compare the normalized envelope with the resonance width; the plots are analytical teaching curves, not recordings.
- τA
- 63.7 ms
- Δf (−3 dB)
- 5.00 Hz
- T60
- 0.44 s