University Physics — Year 1 · Bachelor Year 1

14Mechanical Oscillators: Damping and Resonance

Push a child’s swing at the wrong moment and nothing much happens; push it once per period, gently, and within a dozen swings the child is shrieking at the top of the arc. A washing machine shudders violently at one particular speed of its spin cycle and is quiet above it. A wine glass sings when a wet finger circles its rim at just the right rate, and shatters when a singer holds the same note loudly enough. Every one of these is a resonance: a system with a natural frequency, driven near that frequency, accumulates energy cycle after cycle until friction bleeds it off as fast as it comes in. This chapter takes the mechanical oscillator — free, damped, then driven — and derives the resonance curves whose electrical twins appeared in Chapter 8.

14.1 The harmonic oscillator

Definition 14.1 (Harmonic oscillator)

A system whose displacement $x$ from a stable equilibrium obeys

\ddot x + \omega_0^2\,x = 0

is a harmonic oscillator of natural angular frequency $\omega_0$ (period $T_0 = 2\pi/\omega_0$ ). The solution $x = A\cos(\omega_0t + \varphi)$ has an amplitude and phase fixed by the initial conditions and a period independent of the amplitude (isochronism).

Proposition 14.2 (Where harmonic oscillators come from)

Mass $m$ on a spring $k$ : $\omega_0^2 = k/m$ (Hooke’s law).
Simple pendulum of length $\ell$ , small angles: $\omega_0^2 = g/\ell$ .
Any particle near a stable equilibrium of a potential $E_p$ : $\omega_0^2 = E_p''(x_0)/m$ (Proposition 13.12).
LC circuit: $\omega_0^2 = 1/LC$ , with $q \leftrightarrow x$ , $L \leftrightarrow m$ , $1/C \leftrightarrow k$ .

The energy $E = \tfrac12 m\dot x^2 + \tfrac12 m\omega_0^2x^2 = \tfrac12 m\omega_0^2A^2$ is constant, sloshing between kinetic and potential form twice per period.

Proof. Newton’s second law with $F = -kx$ ; the pendulum equation $\ddot\theta + (g/\ell)\sin\theta = 0$ with $\sin\theta \approx \theta$ ; Taylor’s formula at the minimum of $E_p$ ; the loop law of the LC circuit. Energy: differentiate and use the equation. ∎

Example 14.3 (A bobbing cylinder)

A vertical cylinder of cross-section $S$ floats with a depth $h$ immersed ( $mg = \rho_wgSh$ , buoyancy from Chapter 21). Pushed down by $x$ , it receives an extra upward force $\rho_wgSx$ : $m\ddot x = -\rho_wgSx$ , so $\omega_0^2 = \rho_wgS/m = g/h$ — the same as a pendulum of length $h$ . A buoy immersed by $1\,\mathrm{m}$ bobs with a $2\,\mathrm{s}$ period, whatever its mass.

14.2 The damped oscillator

Proposition 14.4 (Damped oscillator)

With a viscous friction $-\alpha\dot x$ , the equation becomes

\ddot x + \frac{\omega_0}{Q}\dot x + \omega_0^2x = 0, \qquad Q = \frac{\sqrt{km}}{\alpha} = \frac{m\omega_0}{\alpha},

the canonical second-order form of Chapter 7: for $Q > \tfrac12$ a pseudo-periodic decay $x = A\eu^{-t/\tau}\cos(\omega t + \varphi)$ with $\tau = 2Q/\omega_0$ , $\omega = \omega_0\sqrt{1 - 1/4Q^2}$ ; the energy decays as $\eu^{-2t/\tau}$ and

Q = 2\pi\,\frac{\text{energy stored}}{\text{energy lost per period}} \quad (Q \gg 1) .

Proof. Divide $m\ddot x + \alpha\dot x + kx = 0$ by $m$ . For the energy: $E \propto A^2\eu^{-2t/\tau}$ loses the fraction $2T/\tau = 2\pi/Q$ per period when that is small. ∎

Example 14.5 (Ringing times)

A tuning fork at $440\,\mathrm{Hz}$ still sounds ten seconds after being struck: its amplitude fell to about $5\%$ in $3\tau = 10\,\mathrm{s}$ , so $\tau = 3.3\,\mathrm{s}$ and $Q = \omega_0\tau/2 \approx 4600$ . A car’s suspension ( $Q \approx 0.7$ ) stops within one period; a child’s swing ( $Q \approx 20$ ) loses a third of its energy per swing and needs a push every time.

14.3 Forced oscillations and resonance

Theorem 14.6 (Sinusoidally driven oscillator)

Driven by a force $F_0\cos\omega t$ , the damped oscillator

m\ddot x + \alpha\dot x + kx = F_0\cos\omega t

settles, after a transient of duration $\sim\tau$ , into the forced regime $x = X\cos(\omega t + \varphi)$ with, writing $x_r = \omega/\omega_0$ and $X_0 = F_0/k$ the static deflection,

X = \frac{X_0}{\sqrt{(1 - x_r^2)^2 + x_r^2/Q^2}}, \qquad \tan\varphi = -\frac{x_r/Q}{1 - x_r^2}, \quad \varphi \in [-\pi, 0] .

The amplitude $X$ peaks, for $Q > 1/\sqrt2$ , at $\omega_r = \omega_0\sqrt{1 - 1/2Q^2}$ with $X_{\max} \approx QX_0$ ; the velocity amplitude $V = \omega X$ peaks exactly at $\omega_0$ , where $V_{\max} = F_0/\alpha$ , and its resonance curve

\frac{V}{V_{\max}} = \frac{1}{\sqrt{1 + Q^2(x_r - 1/x_r)^2}}

has the bandwidth $\Delta\omega = \omega_0/Q$ . At $\omega_0$ the displacement lags the force by $\pi/2$ and the velocity is in phase with it.

Proof. Complex method (Chapter 8): $\underline X(-\omega^2 + j\omega\omega_0/Q + \omega_0^2) = F_0/m$ , so $\underline X = (F_0/m)/(\omega_0^2 - \omega^2 + j\omega\omega_0/Q)$ ; divide numerator and denominator by $\omega_0^2 = k/m$ . Modulus and argument follow; the maximum of $X$ was located in Proposition 8.10 (same function). Velocity: $\underline V = j\omega\underline X$ gives $V = \omega X$ ; writing $(1 - x_r^2)/x_r = 1/x_r - x_r$ turns the denominator into $\sqrt{1/Q^2 + (1/x_r - x_r)^2}$ , i.e. the RLC form. At $x_r = 1$ , $\underline X = -jQX_0$ : a lag of $\pi/2$ . ∎

Amplitude (left) and phase (right) of the forced oscillator against the driving frequency, for several quality factors. A sharp peak near _0 for large Q, none below Q = 1/√2; the displacement goes from in phase (low ) to opposite (high ), through a π/2 lag at _0 — the sharper the resonance, the more abrupt the switch. — Amplitude (left) and phase (right) of the forced oscillator against the driving frequency, for several quality factors. A sharp peak near $\omega_0$ for large $Q$ , none below $Q = 1/\sqrt2$ ; the displacement goes from in phase (low $\omega$ ) to opposite (high $\omega$ ), through a $\pi/2$ lag at $\omega_0$ — the sharper the resonance, the more abrupt the switch.

Remark 14.7 (Two resonances, one circuit)

The correspondence $x \leftrightarrow q$ , $\dot x \leftrightarrow i$ , $m \leftrightarrow L$ , $\alpha \leftrightarrow R$ , $k \leftrightarrow 1/C$ , $F \leftrightarrow e$ maps the driven oscillator onto the series RLC of Chapter 8: the velocity resonance is the current resonance (always at $\omega_0$ , bandwidth $\omega_0/Q$ ), the amplitude resonance is the charge, i.e. capacitor-voltage, resonance (slightly below $\omega_0$ , absent for low $Q$ ). Power absorbed, $\langle F\dot x\rangle = \tfrac12 F_0V\cos\varphi_v$ , peaks at $\omega_0$ with the value $F_0^2/2\alpha$ , and the half-power points are the velocity bandwidth — exactly as for the RLC.

Example 14.8 (Resonant speeds)

A car’s suspension has $f_0 \approx 1.3\,\mathrm{Hz}$ (Problem 7.1). On a road with corrugations every $\lambda = 10\,\mathrm{m}$ the excitation frequency is $v/\lambda$ : resonance at $v = \lambda f_0 = 13\,\mathrm{m}/\mathrm{s}$ , about $47\,\mathrm{km}/\mathrm{h}$ — the speed at which a washboard track shakes the car hardest, and the reason a damper with $Q \approx 0.7$ is fitted: with $Q = 5$ the body would bounce with five times the road’s amplitude. Above resonance ( $x_r \gg 1$ ) the amplitude falls as $X_0/x_r^2$ : the body “floats” over fast bumps — isolation, the other face of resonance.

14.4 Base excitation: isolation and the seismometer

Proposition 14.9 (Oscillator shaken by its support)

Let the support of a mass–spring–damper move by $x_g(t)$ (the road under a car, the ground under a seismometer), and let $u$ be the displacement of the mass relative to the support and $x = x_g + u$ its absolute displacement (from equilibrium). Then

m\ddot u + \alpha\dot u + ku = -m\ddot x_g ,

and for $x_g = X_g\cos\omega t$ , with $x_r = \omega/\omega_0$ :

\frac{U}{X_g} = \frac{x_r^2}{\sqrt{(1 - x_r^2)^2 + x_r^2/Q^2}}, \qquad \frac{X}{X_g} = \sqrt{\frac{1 + x_r^2/Q^2}{(1 - x_r^2)^2 + x_r^2/Q^2}} .

For $x_r \gg 1$ : $U \to X_g$ (the mass stays still in space — a seismometer reads the ground’s displacement) and $X \to 0$ (the mass is isolated from the shaking); for $x_r \ll 1$ : $U \approx \ddot x_g/\omega_0^2$ (an accelerometer) and $X \approx X_g$ (the mass follows the support).

Proof. The spring and damper act on the relative motion: $m\ddot x = -ku - \alpha\dot u$ , and $\ddot x = \ddot x_g + \ddot u$ . In complex form $\underline U(\omega_0^2 - \omega^2 + j\omega\omega_0/Q) = \omega^2X_g$ , whence $U/X_g$ ; then $\underline X = \underline U + X_g = X_g(1 + jx_r/Q)/(1 - x_r^2 + jx_r/Q)$ . ∎

A mass shaken through its support. Left: relative motion U/X_g — an accelerometer below _0 (slope 2 on log scales), a seismometer above. Right: absolute motion X/X_g — the mass follows the support at low frequency, is isolated from it well above _0; a high Q buys nothing but a dangerous peak. — A mass shaken through its support. Left: relative motion $U/X_g$ — an accelerometer below $\omega_0$ (slope $2$ on log scales), a seismometer above. Right: absolute motion $X/X_g$ — the mass follows the support at low frequency, is isolated from it well above $\omega_0$ ; a high $Q$ buys nothing but a dangerous peak.

Example 14.10 (Why machines sit on soft mounts)

A $50\,\mathrm{Hz}$ motor on rubber mounts tuned to $f_0 = 10\,\mathrm{Hz}$ ( $x_r = 5$ , $Q \approx 2$ ): $X/X_g = \sqrt{(1 + 6.25)/(576 + 6.25)} = 0.11$ — nine tenths of the vibration stays out of the floor. The mounts must be softer than intuition suggests: stiffening them raises $f_0$ toward $50$ Hz and destroys the isolation. The same formula, read the other way, is the car of Example 14.8 on a washboard road.

14.5 Energy: how a resonance builds up

Proposition 14.11 (Energy balance in the forced regime)

In the steady forced regime the average power supplied by the driving force equals the average power dissipated by friction, $\tfrac12\alpha V^2$ . At resonance the stored energy $\tfrac12 mV_{\max}^2$ is $Q/2\pi$ times the energy supplied per period: the oscillator accumulates the drive’s work over about $Q$ cycles before losses match the input, which is also the number of cycles ( $\sim\tau/T$ ) the transient lasts.

Proof. Average the energy theorem over a period: $\langle\dd E/\dd t\rangle = 0$ in steady state, so $\langle F\dot x\rangle = \langle\alpha\dot x^2\rangle = \tfrac12\alpha V^2$ . At resonance $V_{\max} = F_0/\alpha$ , stored energy $\tfrac12 m(F_0/\alpha)^2$ , supplied per period $\tfrac12\alpha(F_0/\alpha)^2T$ ; ratio $m/(\alpha T) = Q/2\pi$ . ∎

Example 14.12 (The swing)

A swing of length $2.5\,\mathrm{m}$ ( $T_0 = 3.2\,\mathrm{s}$ , $Q \approx 20$ ) gets a push of $50\,\mathrm{N}$ over $0.5\,\mathrm{m}$ each period: $25\,\mathrm{J}$ per cycle. It grows until the loss per period, $2\pi E/Q$ , equals $25\,\mathrm{J}$ : $E = 80\,\mathrm{J}$ ; with a $30\,\mathrm{kg}$ child, $mg\ell(1 - \cos\theta_{\max}) = 80\,\mathrm{J}$ gives $\theta_{\max} \approx 27^\circ$ . Pushing at any other rhythm, the work of the pushes would alternate in sign and average to nothing.

Start-up of a resonance (Q = 5, drive switched on at t = 0 at _0): the amplitude grows toward QX_0 with the time constant = 2Q/ _0, i.e. over about Q/π periods — the transient of superposed on the forced regime. — Start-up of a resonance ( $Q = 5$ , drive switched on at $t = 0$ at $\omega_0$ ): the amplitude grows toward $QX_0$ with the time constant $\tau = 2Q/\omega_0$ , i.e. over about $Q/\pi$ periods — the transient of Chapter 7 superposed on the forced regime.

14.6 Exercises

Exercise 14.1 ★

A $0.20\,\mathrm{kg}$ mass on a spring of stiffness $80\,\mathrm{N}/\mathrm{m}$ oscillates with amplitude $5.0\,\mathrm{cm}$ . Angular frequency, frequency, period, maximal speed, total energy.

Solution

Solution of Exercise 14.1.

$\omega_0 = \sqrt{80/0.20} = 20\,\mathrm{rad}/\mathrm{s}$ , $f_0 = 3.2\,\mathrm{Hz}$ , $T_0 = 0.31\,\mathrm{s}$ ; $v_{\max} = A\omega_0 = 1.0\,\mathrm{m}/\mathrm{s}$ ; $E = \tfrac12 kA^2 = 0.10\,\mathrm{J}$ .

Exercise 14.2 ★

Period of a $1.0\,\mathrm{m}$ pendulum on Earth and on the Moon ( $g = 1.62\,\mathrm{m}/\mathrm{s}^{2}$ ). What happens to the mass–spring period on the Moon?

Solution

Solution of Exercise 14.2.

$T = 2\pi\sqrt{\ell/g}$ : $2.0\,\mathrm{s}$ on Earth, $4.9\,\mathrm{s}$ on the Moon. The mass–spring period, $2\pi\sqrt{m/k}$ , does not involve $g$ : unchanged.

Exercise 14.3 ★

An oscillator has $f_0 = 2.0\,\mathrm{Hz}$ and $Q = 50$ . Decay time $\tau$ of the amplitude; time for the energy to halve; number of oscillations before the amplitude is down to $5\%$ .

Solution

Solution of Exercise 14.3.

$\tau = 2Q/\omega_0 = 100/(4\pi) = 8.0\,\mathrm{s}$ ; energy $\propto \eu^{-2t/\tau}$ halves at $t = \tau\ln2/2 = 2.8\,\mathrm{s}$ ; $5\%$ after $3\tau = 24\,\mathrm{s}$ , about $48 \approx Q$ oscillations.

Exercise 14.4 ★

A mass on a spring ( $k = 100\,\mathrm{N}/\mathrm{m}$ , $Q = 10$ ) is driven by a force of amplitude $1.0\,\mathrm{N}$ . Static deflection; amplitude at resonance; amplitude at $\omega = 3\omega_0$ .

Solution

Solution of Exercise 14.4.

$X_0 = F_0/k = 1.0\,\mathrm{cm}$ ; at resonance $\approx QX_0 = 10\,\mathrm{cm}$ ; at $3\omega_0$ : $X_0/\sqrt{64 + 0.09} = 0.13\,\mathrm{cm}$ .

Exercise 14.5 ★★

A car ( $f_0 = 1.3\,\mathrm{Hz}$ , $Q = 0.7$ ) crosses speed bumps $8.0\,\mathrm{m}$ apart. Resonant speed? At $60\,\mathrm{km}/\mathrm{h}$ , what fraction of the bump amplitude does the body feel? Comment.

Solution

Solution of Exercise 14.5.

$v = Lf_0 = 8.0 \times 1.3 = 10\,\mathrm{m}/\mathrm{s}$ ( $37\,\mathrm{km}/\mathrm{h}$ ). At $60\,\mathrm{km}/\mathrm{h}$ : $f = 16.7/8.0 = 2.1\,\mathrm{Hz}$ , $x_r = 1.6$ : $X/X_g = \sqrt{(1 + 5.2)/(2.4 + 5.2)} = 0.90$ — barely isolated; isolation needs $x_r \gg 1$ (at $120\,\mathrm{km}/\mathrm{h}$ , $x_r = 3.2$ : $0.45$ ).

Exercise 14.6 ★★

Explain with the energy balance why a swing must be pushed at its own period; for $Q = 20$ and a push adding $25\,\mathrm{J}$ per cycle, find the steady-state energy, and the amplitude for a $30\,\mathrm{kg}$ child on a $2.5\,\mathrm{m}$ swing.

Solution

Solution of Exercise 14.6.

The work of a push is positive only when it acts along the motion; pushed every period, all pushes add; at any other rhythm they alternate. Steady state: $2\pi E/Q = 25\,\mathrm{J}$ , $E = 80\,\mathrm{J}$ ; $mg\ell(1 - \cos\theta_{\max}) = 80$ : $1 - \cos\theta_{\max} = 0.109$ , $\theta_{\max} = 27^\circ$ .

Exercise 14.7 ★★

Successive maxima of a free oscillation decrease by $10\%$ each period. Find $Q$ and the bandwidth (in hertz) of its velocity resonance if $f_0 = 2.0\,\mathrm{Hz}$ .

Solution

Solution of Exercise 14.7.

$\delta = \ln(1/0.9) = 0.105$ , $Q = \pi/\delta = 30$ ; $\Delta f = f_0/Q = 0.067\,\mathrm{Hz}$ .

Exercise 14.8 ★★

Give the phase of the displacement relative to the force well below resonance, at resonance and well above, and the phase of the velocity at resonance. Why is the velocity resonance the one that matters for power?

Solution

Solution of Exercise 14.8.

Displacement: in phase ( $\varphi = 0$ ) well below, $-\pi/2$ at resonance, $-\pi$ (opposite) well above. Velocity at resonance: in phase with the force. Power is $\vect F\cdot\vect v$ : it is the velocity’s phase, not the displacement’s, that decides how much work the drive does.

Exercise 14.9 ★★

Show that the velocity amplitude $V = \omega X$ is maximal exactly at $\omega_0$ , with $V_{\max} = F_0/\alpha$ , and that $V/V_{\max}$ has the form $1/\sqrt{1 + Q^2(x_r - 1/x_r)^2}$ .

Solution

Solution of Exercise 14.9.

$V = \omega X = \omega(F_0/m)/\sqrt{(\omega_0^2 - \omega^2)^2 + (\omega\omega_0/Q)^2}$ ; divide numerator and denominator by $\omega\omega_0/Q$ : $V = (F_0Q/m\omega_0)/ \sqrt{1 + Q^2(\omega_0/\omega - \omega/\omega_0)^2}$ , and $F_0Q/m\omega_0 = F_0/\alpha$ . The square root is $\geq 1$ , equal to $1$ exactly at $\omega = \omega_0$ .

Exercise 14.10 ★★★

Compute the average power absorbed at resonance for $F_0 = 1.0\,\mathrm{N}$ , $\alpha = 0.050\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ , and the stored energy if $m = 0.20\,\mathrm{kg}$ ; check that their ratio is $Q/\omega_0$ with $\omega_0 = 20\,\mathrm{rad}/\mathrm{s}$ .

Solution

Solution of Exercise 14.10.

$P = F_0^2/2\alpha = 10\,\mathrm{W}$ ; $V_{\max} = F_0/\alpha = 20\,\mathrm{m}/\mathrm{s}$ , $E = \tfrac12 mV_{\max}^2 = 40\,\mathrm{J}$ ; $E/P = 4.0\,\mathrm{s}$ and $Q/\omega_0 = m/\alpha = 4.0\,\mathrm{s}$ .

Exercise 14.11 ★★★

A tuning fork at $440\,\mathrm{Hz}$ is heard for $10\,\mathrm{s}$ (amplitude down to $5\%$ ). Estimate $\tau$ and $Q$ , the bandwidth of its resonance, and comment on why it is a good frequency standard.

Solution

Solution of Exercise 14.11.

$3\tau = 10\,\mathrm{s}$ : $\tau = 3.3\,\mathrm{s}$ , $Q = \omega_0\tau/2 = 2\pi \times 440 \times 3.3/2 \approx 4600$ ; $\Delta f = f_0/Q = 0.1\,\mathrm{Hz}$ . It responds to, and emits, essentially one frequency: a standard good to a few parts in $10^4$ .

Exercise 14.12 ★★★

Derive the bobbing frequency of a floating cylinder (Example 14.3) from Newton’s law and Archimedes’ force $\rho_wgS(h + x)$ ; show $\omega_0^2 = g/h$ ; compute the period for $h = 10\,\mathrm{cm}$ . Why is the result independent of the mass and of the liquid?

Solution

Solution of Exercise 14.12.

$m\ddot x = mg - \rho_wgS(h + x) = -\rho_wgSx$ (since $mg = \rho_wgSh$ ): $\omega_0^2 = \rho_wgS/m = g/h$ ; $T = 2\pi\sqrt{0.10/9.81} = 0.63\,\mathrm{s}$ . Both the mass and the liquid’s density enter only through the equilibrium depth $h$ .

A seismograph: a heavy mass on a soft suspension stays still while the ground moves, and the pen writes the difference — the base-excited oscillator of the weekend problem.

14.7 Problem: The seismograph

Problem 14.1

Weekend problem — a mass hangs from a spring inside a box bolted to the ground; the ground shakes: what does the pen record, why the same instrument measures displacement at high frequency and acceleration at low frequency, and why a long-period seismometer cannot simply be a long spring

A mass $m = 1.0\,\mathrm{kg}$ hangs from a spring of stiffness $k$ inside a rigid frame fixed to the ground; a damper exerts $-\alpha\dot u$ , where $u$ is the displacement of the mass relative to the frame, measured from its equilibrium position. The ground, hence the frame, moves vertically by $x_g(t)$ . The frame of the distant stars is inertial.

Part I — The equation.

Write the absolute position of the mass as $x = x_g + u + \text{const}$ , list the forces on the mass (in the inertial frame), and show that

$m\ddot u + \alpha\dot u + ku = -m\ddot x_g .$
Put it in canonical form; identify $\omega_0$ and $Q$ .
The instrument is built with $f_0 = 0.50\,\mathrm{Hz}$ and $Q = 0.70$ : compute $k$ and $\alpha$ .
A constant ground acceleration $a_g$ (the frame tilting, say) gives what steady $u$ ? Interpret.
What does the pen attached to the mass, writing on a drum fixed to the frame, record?

Part II — Harmonic ground motion. The ground moves as $x_g = X_g\cos\omega t$ ; write $x_r = \omega/\omega_0$ .

Using complex amplitudes, show that $\underline U = X_g\,x_r^2/ (1 - x_r^2 + jx_r/Q)$ .
Deduce $U/X_g$ as a function of $x_r$ and $Q$ .
Limit $x_r \gg 1$ : show $U \to X_g$ and explain physically what the mass does (what is the instrument measuring?).
Limit $x_r \ll 1$ : show $U \approx x_r^2X_g = a_g/\omega_0^2$ with $a_g$ the ground’s acceleration amplitude: what is the instrument now?
Value at $x_r = 1$ ; why is $Q \approx 0.7$ a sensible choice?
Sketch $U/X_g$ against $x_r$ (log scales) for $Q = 0.7$ and $Q = 5$ .
Phase of $u$ relative to $x_g$ at high frequency; interpret.

Part III — Designing for earthquakes. Distant earthquakes produce surface waves at $0.05\,$ to $0.1\,\mathrm{Hz}$ with millimeter amplitudes; local tremors reach $5\,$ to $10\,\mathrm{Hz}$ .

To record $0.05\,\mathrm{Hz}$ waves as displacements one wants $f_0 \ll 0.05\,\mathrm{Hz}$ , say $0.02\,\mathrm{Hz}$ . What static stretch would the spring have under the mass’s weight? Comment.
A simple pendulum of the same frequency: what length?
Real long-period instruments use a nearly horizontal “garden-gate” pendulum, or electronic force feedback. Explain in one sentence what each achieves.
For the $0.50\,\mathrm{Hz}$ instrument, compute the static stretch: feasible?
With it, a $0.10\,\mathrm{Hz}$ wave of amplitude $1.0\,\mathrm{mm}$ gives what $U$ ? In which regime is the instrument?
A $5.0\,\mathrm{Hz}$ local tremor of amplitude $1.0\,\mathrm{mm}$ : $U$ ? Regime?

Part IV — The accelerometer in a phone. A micro-machined mass on silicon springs has $f_0 = 1.0\,\mathrm{kHz}$ and $Q = 0.7$ .

A $10\,\mathrm{Hz}$ shake of acceleration amplitude $g$ : compute $U$ . How is such a displacement read?
Up to what frequency is the reading within $10\%$ of $a_g/\omega_0^2$ ?
What time constant $\tau = 2Q/\omega_0$ governs its response to a jolt, and why does that matter for detecting a fall?
The seismometer’s $\tau$ : value and consequence after a sudden ground step.
Why is critical (or near-critical) damping chosen for both instruments, rather than a high $Q$ ?
The phone is shaken at exactly $1.0\,\mathrm{kHz}$ : by what factor is the reading off?
Summarize: one device, two regimes, decided by which single comparison.

Solution

Solution of Problem 14.1.

1. Forces: weight (balanced by the static stretch), spring $-ku$ , damper $-\alpha\dot u$ ; $m\ddot x = -ku - \alpha\dot u$ with $\ddot x = \ddot x_g + \ddot u$ .

2. $\ddot u + (\omega_0/Q)\dot u + \omega_0^2u = -\ddot x_g$ , $\omega_0 = \sqrt{k/m}$ , $Q = \sqrt{km}/\alpha$ .

3. $k = m\omega_0^2 = \pi^2 = 9.9\,\mathrm{N}/\mathrm{m}$ ; $\alpha = m\omega_0/Q = 3.14/0.70 = 4.5\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ .

4. $u = -a_g/\omega_0^2$ : the mass sags by a fixed amount — the instrument is an accelerometer at zero frequency (and a tilt meter).

5. The relative displacement $u(t)$ .

6. $-\omega^2\underline U + j\omega(\omega_0/Q)\underline U + \omega_0^2\underline U = \omega^2X_g$ ; divide by $\omega_0^2$ .

7. $U/X_g = x_r^2/\sqrt{(1 - x_r^2)^2 + x_r^2/Q^2}$ .

8. $U \to X_g$ : the mass does not move in the inertial frame (its inertia keeps it still while the spring is too soft to drag it); the frame moves around it, and the pen records the ground displacement — a seismometer.

9. $U \approx x_r^2X_g = \omega^2X_g/\omega_0^2 = a_g/\omega_0^2$ : an accelerometer.

10. $U = QX_g = 0.7X_g$ : no peak; the two regimes join smoothly and the instrument rings not.

11. For $Q = 0.7$ : a line of slope $2$ rising to $1$ near $x_r = 1$ , then flat at $1$ ; for $Q = 5$ : the same with a peak of height $5$ at $x_r = 1$ .

12. $\underline U \to -X_g$ : $u = -x_g$ , the mass still while the frame moves.

13. $\Delta\ell = g/\omega_0^2 = 9.81/(2\pi \times 0.02)^2 = 620\,\mathrm{m}$ : absurd.

14. $\ell = g/\omega_0^2 = 620\,\mathrm{m}$ .

15. Garden gate: the pendulum swings about a nearly vertical axis, so only a tiny fraction of $g$ restores it and a short arm gives a long period. Force feedback: a coil holds the mass fixed, and the current needed is proportional to the ground acceleration — the stiffness is electronic and adjustable.

16. $\Delta\ell = 9.81/\pi^2 = 0.99\,\mathrm{m}$ : feasible.

17. $x_r = 0.2$ : $U/X_g = 0.04/\sqrt{0.92 + 0.08} = 0.040$ : $40\,\text{µ}\mathrm{m}$ — accelerometer regime, needs amplification (optical lever or electronics).

18. $x_r = 10$ : $100/\sqrt{9801 + 204} \approx 1.00$ : $1.0\,\mathrm{mm}$ , faithful displacement record.

19. $\omega_0 = 6.3 \times 10^{3}\,\mathrm{rad}/\mathrm{s}$ : $U = g/\omega_0^2 = 2.5 \times 10^{-7}\,\mathrm{m}$ — read as the change of a capacitor gap.

20. $U/(a_g/\omega_0^2) = 1/\sqrt{(1 - x_r^2)^2 + x_r^2/Q^2}$ ; with $Q = 0.7$ it decreases monotonically and stays above $0.9$ up to $x_r \approx 0.7$ : about $700\,\mathrm{Hz}$ .

21. $\tau = 2Q/\omega_0 = 0.22\,\mathrm{ms}$ : it tracks a fall (tens of milliseconds) without lag.

22. $\tau = 1.4/3.14 = 0.45\,\mathrm{s}$ : after a ground step the record settles in half a second with no ringing.

23. A high $Q$ would make the instrument ring at $f_0$ after every jolt (recording itself rather than the ground) and peak near $f_0$ ; $Q \approx 0.7$ is flat and quick.

24. $U = QX_g = 0.7X_g$ against the ideal $X_g$ (or $a_g/\omega_0^2 = X_g$ ): $30\%$ low.

25. Compare the signal frequency with $f_0$ : well above, the mass is a fixed reference and $u$ is the displacement; well below, $u$ is the acceleration over $\omega_0^2$ .

Terms defined in this chapter

Harmonic oscillator

See all 393 terms in the glossary