# Resonance Curve

Wave · y = f(x, t)

`y = (1/(sqrt((ω₀² − x²)² +(γ · x)²)))`

[Open in the app](https://www.wavelace.com/app#p=130) · [This page](https://www.wavelace.com/presets/resonance-curve)

### What it draws

This is the steady amplitude of a damped oscillator against the frequency it is driven at, which is `x`. `ω₀` is the oscillator's own natural frequency, opening at 3, and `γ` is how strongly it is damped, opening at 0.5. There is no `t` anywhere in the expression, so the curve stands still while the clock runs.

The denominator is the distance from the origin to the point `(ω₀² − x², γ·x)`. The first part vanishes when the drive matches the oscillator, at `x = ω₀`, and only the damping term is left to keep the amplitude finite there. At `x = 0` the value is `1/ω₀² = 1/9 ≈ 0.111` whatever the damping, since a steady drive just pushes the mass to one side.

### Why the peak

Maximising the response means minimising `(ω₀² − x²)² + (γ·x)²`, which puts the peak at `√(ω₀² − γ²/2) ≈ 2.979` and not quite at `ω₀` itself. Its height is `1/(γ·√(ω₀² − γ²/4)) ≈ 0.669`, six times the value at zero frequency.

Far from resonance the two squares are dominated by `x⁴`, so the curve dies away as `1/x²`. At `x = 10` it is down to 0.011, a factor of 61 below the peak. Only `x²` appears, so the plot is even: the mirrored hump on the left is the same resonance at a negative driving frequency.

### How sharp the peak is

Damping moves the picture a long way. At the opening value the peak holds half its height over a band 0.44 wide either side; at `γ = 0.1` it reaches 3.33 and that band narrows to 0.087. Raise the damping instead and the resonance drains away, the peak down to 0.128 by `γ = 3`, barely above the 0.111 a steady push gives.

The peak position is real only while `γ` is under `ω₀·√2 ≈ 4.243`, and past that the only maximum sits at `x = 0`. The threshold is exact and all but invisible: at `γ = 4` the peak stands 0.62% above the value at zero frequency, a dimple no plot resolves. Past it no drive makes the oscillator swing further than a steady push would.

### Try

- Drag `γ` down to 0.2: the peak climbs to 1.67 and narrows to match, a sharply tuned oscillator.
- Drag `γ` to 4.2: two peaks are still there on paper, at `±0.424`, but they stand 0.02% above the centre, so the plot reads as one low hump.
- Drag `ω₀` to 6: the peak moves out to 5.99 and drops to 0.334, since a stiffer oscillator gives less for the same push.
- Change the `1` on top to `x`, the velocity response: its peak sits exactly at `x = ω₀` for every damping, at a height of `1/γ = 2`.
- Raise `Span of x` to 20: the tail is then long enough to read the `1/x²` fall.

### Read more

- [Resonance](https://en.wikipedia.org/wiki/Resonance)
- [Harmonic oscillator](https://en.wikipedia.org/wiki/Harmonic_oscillator)
- [Q factor](https://en.wikipedia.org/wiki/Q_factor)
- [Damping](https://en.wikipedia.org/wiki/Damping)