Part 1 · The elementary oscillator

1.2 The free undamped oscillator

Source: Carboni’s notes for lesson 1, Oscillatore elementare, pp. 3–5

Objectives

Recipe. Solve $M \ddot{U} + K U = 0$ from two initial conditions: characteristic equation, then amplitude and phase.
Vocabulary. Natural circular frequency $ω$ , natural frequency $f$ , natural period $T$ .
Picture. A sinusoid whose period depends only on $K$ and $M$ , and whose height depends only on how it was started.
Recipe. Weigh a molecule from the shift of a MEMS resonator’s frequency.

Pull the mass aside, let it go, and nothing else acts on it. The equation of motion is the one of Definition 1.1.5, $M \ddot{U} + K U = 0, {U (0) = U_{0} \dot{U} (0) = V_{0}$ The equation is of second order in time, so it takes two initial conditions to fix the motion: the starting position $U_{0}$ and the starting velocity $V_{0}$ (both numbers).

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Figure 1.2.1 The free undamped response. Drag the violet point to change the starting position

U_{0}

and the teal one to change the starting slope, which is the starting velocity

V_{0}

. The amplitude

A

(dashed) and the phase change; the period does not. Change

K

and

M

in the controls to change the period.

Solving the equation

Recipe 1.2.1solving the free undamped oscillatornotes L1, p. 3

Try an exponential. $U (t) = c_{1} e^{μ t} + c.c.$ , where c.c. is the complex conjugate of the term before it, added so that $U$ is real. $μ$ is a number to be found.
Substitute. Since $\ddot{U} = c_{1} μ^{2} e^{μ t} + c.c.$ , $c_{1} μ^{2} M e^{μ t} + K c_{1} e^{μ t} = c_{1} (M μ^{2} + K) e^{μ t} = 0 \forall t .$
Characteristic equation. $e^{μ t}$ is never zero, and $c_{1} = 0$ is the mass sitting still. So the bracket vanishes: $M μ^{2} + K = 0 ⟹ μ = \pm - \frac{K}{M} = \pm i \frac{K}{M} .$
Name the root. Write $μ = \pm iω$ with $ω = K / M$ . Then $U (t) = c_{1} e^{iω t} + c_{2} e^{- iω t}$ .
Back to real functions with Euler’s formulas $cos ω t = \frac{e ^{iω t} + e ^{- iω t}}{2}, sin ω t = \frac{e ^{iω t} - e ^{- iω t}}{2 i} :$ $U (t) = a sin (ω t) + b cos (ω t) = A cos (ω t - ϕ), A = a^{2} + b^{2}, ϕ = arctan \frac{a}{b} .$
Initial conditions (beyond the lectures): $U (0) = b = U_{0}$ and $\dot{U} (0) = aω = V_{0}$ .

So the motion that starts from $U_{0}$ with velocity $V_{0}$ is $U (t) = \frac{V _{0}}{ω} sin ω t + U_{0} cos ω t .$

Why the two forms agree

Expand the cosine of a difference: $A cos (ω t - ϕ) = A cos ϕ cos ω t + A sin ϕ sin ω t .$ Matching with $a sin ω t + b cos ω t$ term by term gives $a = A sin ϕ$ and $b = A cos ϕ$ . Squaring and adding: $a^{2} + b^{2} = A^{2} (sin^{2} ϕ + cos^{2} ϕ) = A^{2}$ . Dividing: $a / b = tan ϕ$ .

Warningslip in the notes

The notes write $ϕ = arctan (b / a)$ , here and again for the damped oscillator (p. 8). From $a = A sin ϕ$ and $b = A cos ϕ$ , $tan ϕ = a / b$ , so $ϕ = arctan (a / b)$ . With $arctan (b / a)$ the formula would have to be $A sin (ω t + ϕ)$ instead.

Why an exponential is the right guess

A function whose second derivative is a multiple of itself is what the equation asks for: $\ddot{U} = - (K / M) U$ . Exponentials are the functions whose derivatives are multiples of themselves, $\frac{d}{d t} e^{μ t} = μ e^{μ t}$ , so trying $e^{μ t}$ turns the differential equation into an algebraic one for $μ$ . The same trick solves the damped oscillator in §1.3 The damped oscillator.

Frequency and period

Definition 1.2.2natural frequencyEssentialnotes L1, p. 4

For the oscillator $M \ddot{U} + K U = 0$ :

$ω = K / M$ , in rad/s, is the natural circular frequency;
$f = \frac{ω}{2 π}$ , in Hz, is the natural frequency (frequenza propria di vibrare);
$T = \frac{1}{f} = \frac{2 π}{ω}$ , in s, is the natural period (periodo proprio), the time between two maxima.

$U (t)$ oscillates between $- A$ and $A$ with period $T$ . With no energy dissipation (damping), the oscillation goes on for ever.

Key observation 1.2.3what natural meansbeyond the lectures

$ω$ depends only on $K$ and $M$ : not on the initial conditions, not on the amplitude. That is what natural means: it belongs to the system, and the start only sets $A$ and $ϕ$ . A stiffer spring raises it, a heavier mass lowers it. In Figure 1.2.1, dragging the start never moves the dotted period lines.

Example 1.2.4numbersbeyond the lectures

$M = 0.5$ kg, $K = 200$ N/m, started at $U_{0} = 10$ mm with $V_{0} = 0.1$ m/s.

$ω = 200/0.5 = 20$ rad/s, $f = 20/2 π = 3.183$ Hz, $T = 0.3142$ s.
$b = U_{0} = 0.01$ m, $a = V_{0} / ω = 0.005$ m.
$A = a^{2} + b^{2} = 11.18$ mm, $ϕ = arctan (a / b) = arctan 0.5 = 0.4636$ rad $= 26.57°$ .

Mass detection with MEMS

An engineering use of the natural frequency: weighing a single molecule, or detecting a gas at a very low concentration. A MEMS resonator (a micro-electro-mechanical system) behaves like a spring $K$ and a mass $M$ . When a small mass $m$ (the molecule) lands on it, the mass grows and the frequency drops.

Recipe 1.2.5mass detection from a frequency shiftnotes L1, pp. 4–5

Reference frequency, before: $ω_{0} = K / M$ .
Shifted frequency, after the mass $m$ lands: $\overset{ω}{ˉ} = K / (M + m)$ .
Measure the shift $Δ ω = ω_{0} - \overset{ω}{ˉ}$ . In practice it is read from the Fourier transform of the free oscillation: the spectrum has a peak at the natural frequency, and the peak moves left, from $ω_{0}$ to $\overset{ω}{ˉ}$ .
Solve for $m$ . For $m ≪ M$ (beyond the lectures): $m ≃ 2 M Δ ω / ω_{0}$ .

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Figure 1.2.2 The spectrum of the resonator’s free oscillation, before (grey, dashed) and after (violet) a mass

m

lands on it. Drag the violet peak or change

m / M

: the shift grows in proportion to the added mass.

Proposition 1.2.6small added massbeyond the lectures

For $m ≪ M$ , $Δ ω ≃ ω_{0} \frac{m}{2 M}, m ≃ 2 M \frac{Δ ω}{ω _{0}} .$

Proof

$\overset{ω}{ˉ} = K / M (1 + m / M)^{- 1/2} = ω_{0} (1 + m / M)^{- 1/2}$ . For a small $x$ , $(1 + x)^{- 1/2} = 1 - \frac{1}{2} x + O (x^{2})$ , so $\overset{ω}{ˉ} ≃ ω_{0} (1 - m /2 M)$ and $Δ ω = ω_{0} - \overset{ω}{ˉ} ≃ ω_{0} m /2 M$ .

The shift is proportional to the added mass, and larger for a light, stiff resonator (high $ω_{0}$ , small $M$ ): that is why these sensors are made small.

A nonlinear oscillator, for example one with a cubic spring (Definition 1.1.2), $M \ddot{U} + K U + \frac{K _{3}}{3 !} U^{3} = 0,$ has other features in its response that can be used in the same applications.

What the cubic term changesbeyond the lectures

The notes do not say which features. The best known: the frequency of a nonlinear oscillator depends on its amplitude. For a small $K_{3}$ , a first-order calculation (the Lindstedt–Poincaré method) gives $ω_{N L} ≃ ω (1 + \frac{K _{3} A ^{2}}{16 K}) .$ A hardening spring ( $K_{3} > 0$ ) raises the frequency as the amplitude grows, a softening one lowers it. A sensor can track this amplitude–frequency relation, or the jumps between branches it causes when the oscillator is forced.

Exercises

Exercise 1.2.1

A mass $M = 2$ kg on a spring $K = 800$ N/m is released from rest at $U_{0} = 5$ mm. Find $ω$ , $f$ , $T$ , $A$ and $ϕ$ , and write $U (t)$ .

Solution

$ω = 400 = 20$ rad/s, $f = 3.183$ Hz, $T = 0.3142$ s.
From rest, $V_{0} = 0$ : $a = 0$ , $b = U_{0}$ . So $A = 5$ mm and $ϕ = arctan 0 = 0$ .
$U (t) = 5 cos (20 t)$ mm.

Exercise 1.2.2

Which of these changes the natural period of a mass on a spring: doubling the starting displacement, doubling the mass, doubling the stiffness? By how much?

Solution

Doubling $U_{0}$ : no change. $T = 2 π M / K$ does not contain the initial conditions.
Doubling $M$ : $T$ grows by $2$ .
Doubling $K$ : $T$ shrinks by $2$ .

Exercise 1.2.3

A MEMS resonator has $K = 10$ N/m and $M = 1 0^{- 12}$ kg. A particle of $m = 1 0^{- 15}$ kg lands on it. Find $f_{0}$ and the frequency shift $Δ f$ , exactly and with Proposition 1.2.6.

Solution

$ω_{0} = 10/1 0^{- 12} = 3.162 \times 1 0^{6}$ rad/s, $f_{0} = 503.3$ kHz.
Exact: $Δ ω = ω_{0} - K / (M + m) = 1580.0$ rad/s, so $Δ f = 251.5$ Hz.
Approximate: $Δ ω ≃ ω_{0} m /2 M = 1581.1$ rad/s, within 0.1% since $m / M = 1 0^{- 3}$ .
A thousandth of the mass shifts the frequency by half a thousandth.

Exercise 1.2.4

The spectrum of a resonator with $f_{0} = 1$ MHz and $M = 5 \times 1 0^{- 13}$ kg shows its peak moved down by 40 Hz. Estimate the added mass.

Solution

$m ≃ 2 M Δ f / f_{0} = 2 \cdot 5 \times 1 0^{- 13} \cdot 40/1 0^{6} = 4 \times 1 0^{- 17}$ kg (40 attograms). The ratio of circular frequencies equals the ratio of frequencies, so $Δ ω / ω_{0} = Δ f / f_{0}$ .