---
title: "Mechanical Oscillators and the Measurement of Time"
book: "High School Physics"
subject: physics
language: en
chapter: 28
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time
---

# Chapter 28 — Mechanical Oscillators and the Measurement of Time

A grandfather clock ticks: each tick is a brass pendulum crossing the vertical, repeated fifteen million times since New Year. A quartz sliver in your watch flexes $32\,768$ times a second; a cesium [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) hums nine billion times faster. Everything that swings can keep time: this chapter follows the swinging, from Huygens to the [satellites](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite).

## 28.1 Free oscillations

**Definition 28.1 (Oscillator, period, frequency, amplitude).**

A *mechanical oscillator* is a system that, displaced from a stable *equilibrium position* and released, swings back and forth around it: these are *free oscillations*. The motion repeats after the *period* $T_0$ ($\mathrm{s}$); the *frequency* $f_0 = 1/T_0$ ($\mathrm{Hz}$, [Chapter 20](https://one-course.com/books/physics/2/en/chapter/20-mechanical-waves#ch-g12-mechanical-waves)); the *amplitude* is the maximum displacement.

**Example 28.2 (A zoo of oscillators).**

A playground swing: $T_0 \approx 3\,\mathrm{s}$. A guitar’s A string: $f_0 = 110\,\mathrm{Hz}$. A quartz crystal: $32\,768\,\mathrm{Hz}$. The cesium oscillation defining the second: about $9.2 \times 10^{9}\,\mathrm{Hz}$. Ten orders of magnitude, one job: counting.

## 28.2 The simple pendulum

**Definition 28.3 (Simple pendulum).**

A *simple pendulum*: a small bob of mass $m$ on an inextensible wire of length $L$ and negligible mass, swinging in a vertical plane. Of the two [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the bob ([Chapter 25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)), the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is perpendicular to the motion, while the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight)’s component along the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) points back toward the lowest point: the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) is the *restoring [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force)*, opposing the displacement.

![Forces on the bob: the tension T is perpendicular to the motion; the tangential part P_t of the weight always points back toward the vertical — the restoring force.](https://one-course.com/images/onecourse/chapters/physics-2/g12-oscillators-and-time/fig-4439374d74c4.svg)

*[Forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the bob: the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect T$ is perpendicular to the motion; the tangential part $\vect{P_t}$ of the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) always points back toward the vertical — the [restoring force](#def-g12-oscillators-and-time-pendulum).*

**Proposition 28.4 (Period of the simple pendulum).**

For small [amplitudes](#def-g12-oscillators-and-time-oscillator) (below about $15^\circ$), the [period](#def-g12-oscillators-and-time-oscillator) is

$$
T_0 = 2\pi \sqrt{\frac{L}{g}}, \qquad g = 9.81\,\mathrm{m}/\mathrm{s}^{2}.
$$

It depends neither on the mass of the bob nor — while the angle stays small — on the [amplitude](#def-g12-oscillators-and-time-oscillator): small swings are *isochronous*, all taking the same time.

**Proof.** *Admitted at this level.* ∎

**Remark 28.5 (Dimensional analysis, and where the 2π2\pi2π lives).**

The formula can be guessed. From $m$ ($\mathrm{kg}$), $L$ ($\mathrm{m}$) and $g$ ($\mathrm{m}/\mathrm{s}^{2}$), the only combination with the [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of a time is $\sqrt{L/g}$ — and the mass *cannot* enter, no other datum carrying a [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) to cancel it. Dimensions cannot give the pure number in front: the $2\pi$ comes from solving the equation of motion, done in the Year 1 volume. Nor is the small-angle condition optional: at $30^\circ$ the true [period](#def-g12-oscillators-and-time-oscillator) is $2\%$ longer, and isochronism fails.

**Example 28.6 (The seconds pendulum).**

Which length beats the second, $T_0 = 2.00\,\mathrm{s}$, one tick each way? Solving, $L = g\,T_0^2/4\pi^2 = 9.81 \times 2.00^2/4\pi^2 \approx
0.994\,\mathrm{m}$. Nearly a meter: pendulum clocks are tall because seconds are long.

## 28.3 The spring–mass oscillator

**Definition 28.7 (Restoring force of a spring).**

A glider of mass $m$ slides on a horizontal frictionless rail, tied to a spring of *stiffness* $k$ ($\mathrm{N}/\mathrm{m}$). Let $x$ be its displacement from equilibrium. Stretched or compressed, the spring pulls or pushes the glider back with the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $F = -kx$: proportional to the displacement and opposed to it — a [restoring force](#def-g12-oscillators-and-time-pendulum) again, now supplied by elasticity.

**Proposition 28.8 (Equation of motion and its solution).**

[Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) along the rail ([Chapter 25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)) gives $m \frac{d^2x}{dt^2} = -kx$, i.e. $\frac{d^2x}{dt^2} =
-\frac{k}{m}\,x$. Whatever the constants $A$ and $\varphi$,

$$
x(t) = A \cos\!\left(\frac{2\pi t}{T_0} + \varphi\right),
\qquad T_0 = 2\pi\sqrt{\frac{m}{k}},
$$

satisfies this equation: oscillation with [period](#def-g12-oscillators-and-time-oscillator) $T_0$, independent of the [amplitude](#def-g12-oscillators-and-time-oscillator).

**Proof.** Differentiating, $\frac{dx}{dt} = -\frac{2\pi}{T_0}A \sin(\frac{2\pi
t}{T_0} + \varphi)$; again, $\frac{d^2x}{dt^2} = -(\frac{2\pi}{T_0})^2
A\cos(\frac{2\pi t}{T_0} + \varphi) = -(\frac{2\pi}{T_0})^2\,x(t)$, which is $-(k/m)\,x$ exactly when $(2\pi/T_0)^2 = k/m$. ∎

**Remark 28.9 (All the motions there are).**

That *every* motion of the glider is one of these functions is admitted, proved in the Year 1 volume. And no $g$ in $2\pi\sqrt{m/k}$: stiffer swings faster, heavier slower, the same on the Moon.

**Definition 28.10 (Amplitude and phase).**

In $x(t) = A\cos(2\pi t/T_0 + \varphi)$ the positive constant $A$ is the [amplitude](#def-g12-oscillators-and-time-oscillator) — the extremes are $x = \pm A$ — and $2\pi t/T_0 +
\varphi$ is the *phase*, growing by $2\pi$ each [period](#def-g12-oscillators-and-time-oscillator); the *initial phase* $\varphi$ records the start: released at rest from $x = A$, $\varphi = 0$. For instance $m = 0.20\,\mathrm{kg}$ on $k = 20\,\mathrm{N}/\mathrm{m}$ gives $T_0 =
2\pi\sqrt{0.20/20} = 0.63\,\mathrm{s}$; pulled to $5.0\,\mathrm{cm}$ and released, $x(t) = 5.0\cos(2\pi t/0.63)$ in centimeters.

## 28.4 Energy, damping, resonance

**Proposition 28.11 (Energy of the oscillator).**

A spring stretched by $x$ stores the elastic potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) $E_p =
\tfrac12 kx^2$ (justified in [Chapter 29](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#ch-g12-work-and-energy)). During frictionless oscillation the [mechanical energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em)

$$
E_m = E_k + E_p = \tfrac12 m v^2 + \tfrac12 k x^2 = \tfrac12 k A^2
$$

is constant: all elastic at the extremes, all kinetic at equilibrium. The pendulum plays the same duet between $E_k$ and gravitational $E_p$ ([Chapter 18](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#ch-g11-mechanical-energy)).

**Proof.** With $v = \frac{dx}{dt}$, the derivative of $E_m$ is $mv\frac{dv}{dt}
+ kx\frac{dx}{dt} = v\,(m\frac{d^2x}{dt^2} + kx) = 0$ by the equation of motion; the value $\tfrac12 kA^2$ is read at an extreme. ∎

![One period, released from x = A: the energy sloshes twice from elastic to kinetic and back, the sum pinned at 1/2 kA2.](https://one-course.com/images/onecourse/chapters/physics-2/g12-oscillators-and-time/fig-415095a0529f.svg)

*One [period](#def-g12-oscillators-and-time-oscillator), released from $x = A$: the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) sloshes twice from elastic to kinetic and back, the sum pinned at $\tfrac12 kA^2$.*

**Definition 28.12 (Damped oscillations).**

[Friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) drains [mechanical energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em) from every real [oscillator](#def-g12-oscillators-and-time-oscillator): its [free oscillations](#def-g12-oscillators-and-time-oscillator) are *damped*. Under light damping the system still oscillates, with a dying [amplitude](#def-g12-oscillators-and-time-oscillator) but a nearly unchanged repeat time, the *[pseudo-period](#def-g12-oscillators-and-time-oscillator)*, close to $T_0$; under heavy damping (honey) it creeps back to equilibrium without ever overshooting; the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), proportional to [amplitude](#def-g12-oscillators-and-time-oscillator) squared, decays into heat.

![Undamped and lightly damped oscillations: nearly the same period (2.0\, s here), the damped amplitude dying inside a shrinking exponential envelope (dashed).](https://one-course.com/images/onecourse/chapters/physics-2/g12-oscillators-and-time/fig-a341419a3639.svg)

*Undamped and lightly [damped](#def-g12-oscillators-and-time-damping) oscillations: nearly the same [period](#def-g12-oscillators-and-time-oscillator) ($2.0\,\mathrm{s}$ here), the [damped](#def-g12-oscillators-and-time-damping) [amplitude](#def-g12-oscillators-and-time-oscillator) dying inside a shrinking exponential envelope (dashed).*

**Definition 28.13 (Forced oscillations and resonance).**

Push an [oscillator](#def-g12-oscillators-and-time-oscillator) [periodically](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-period) — *forced oscillations* — and after a brief transient it oscillates at the *driving* [frequency](#def-g12-oscillators-and-time-oscillator) $f$, not its own. Its [amplitude](#def-g12-oscillators-and-time-oscillator) peaks when $f$ nears the *natural [frequency](#def-g12-oscillators-and-time-oscillator)* $f_0$ of the free [oscillator](#def-g12-oscillators-and-time-oscillator): this is *resonance*, each push arriving in step with the motion, feeding in [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) every cycle; the lighter the [damping](#def-g12-oscillators-and-time-damping), the taller and sharper the peak.

![Resonance curve: forced amplitude against driving frequency. Near f = f_0 the response towers, higher the lighter the damping.](https://one-course.com/images/onecourse/chapters/physics-2/g12-oscillators-and-time/fig-178449686a50.svg)

*[Resonance](#def-g12-oscillators-and-time-resonance) curve: forced [amplitude](#def-g12-oscillators-and-time-oscillator) against driving [frequency](#def-g12-oscillators-and-time-oscillator). Near $f = f_0$ the response towers, higher the lighter the [damping](#def-g12-oscillators-and-time-damping).*

**Example 28.14 (Resonance for good and ill).**

A parent pushing a swing pushes at its own [frequency](#def-g12-oscillators-and-time-oscillator) — [resonance](#def-g12-oscillators-and-time-resonance) put to [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work). In $2000$ the new Millennium Bridge in London swayed alarmingly when the crowd’s steps locked onto a [natural frequency](#def-g12-oscillators-and-time-resonance) near $1\,\mathrm{Hz}$; it closed two years for dampers. Every quartz watch keeps its crystal singing by driving it at [resonance](#def-g12-oscillators-and-time-resonance), $32\,768\,\mathrm{Hz}$.

## 28.5 Measuring time

**Method 28.15 (Timing a period).**

Never time one oscillation: start the stopwatch as the bob crosses the vertical (fastest, easiest to judge), count $N = 50$ oscillations, divide by $N$ — a $0.2\,\mathrm{s}$ reaction error becomes $4\,\mathrm{ms}$ on $T_0$. Repeat, and compare runs.

**Example 28.16 (Three centuries of tick).**

Huygens built the first pendulum clock in $1657$: isochronism took clocks from a quarter hour of drift per day to seconds, and compensated pendulums to fractions of a second per day. The quartz wristwatch ($1969$) swapped the rod for a crystal flexing at $32\,768\,\mathrm{Hz}$: tenths of a second per month. Since $1967$ the second itself is *defined* by an atomic oscillation: the duration of $9\,192\,631\,770$ [periods](#def-g12-oscillators-and-time-oscillator) of the microwave radiation of an internal transition of cesium-133 ([Chapter 34](https://one-course.com/books/physics/2/en/chapter/34-the-quantum-world-photons-and-energy-levels#ch-g12-quantum-world)). The best cesium clocks drift one second in a hundred million years.

**Remark 28.17 (Why chase the nanosecond).**

A GPS receiver finds its position by timing radio [signals](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal) from [satellites](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite); light travels $30\,\mathrm{cm}$ per nanosecond, so a clock error of one microsecond misplaces you by $300\,\mathrm{m}$. [Satellites](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) therefore carry atomic clocks — and at that precision the rate of a clock depends on how fast it moves and how high it sits, an effect we meet in [Chapter 35](https://one-course.com/books/physics/2/en/chapter/35-special-relativity-time-dilation#ch-g12-special-relativity).

## 28.6 Exercises

**Exercise 28.1 ★.**

(a) Compute the [period](#def-g12-oscillators-and-time-oscillator) and [frequency](#def-g12-oscillators-and-time-oscillator) of a $1.00\,\mathrm{m}$ [simple pendulum](#def-g12-oscillators-and-time-pendulum). (b) The length is quadrupled: what happens to the [period](#def-g12-oscillators-and-time-oscillator)?

**Solution of Exercise 28.1.**

(a) $T_0 = 2\pi\sqrt{1.00/9.81} = 2.01\,\mathrm{s}$; $f_0 = 1/T_0 = 0.50\,\mathrm{Hz}$. (b) $T_0 \propto \sqrt{L}$: the [period](#def-g12-oscillators-and-time-oscillator) doubles.

**Exercise 28.2 ★.**

A resting heart beats $72$ times per minute; a dragonfly’s wing at $30\,\mathrm{Hz}$. Give the heart’s [frequency](#def-g12-oscillators-and-time-oscillator) and [period](#def-g12-oscillators-and-time-oscillator), the wing’s [period](#def-g12-oscillators-and-time-oscillator); which chapter word names the maximum chest rise per beat?

**Solution of Exercise 28.2.**

Heart: $f = 72/60 = 1.2\,\mathrm{Hz}$, $T = 0.83\,\mathrm{s}$. Wing: $T = 1/30 = 33\,\mathrm{ms}$. The chest rise is the *[amplitude](#def-g12-oscillators-and-time-oscillator)*.

**Exercise 28.3 ★.**

A $0.10\,\mathrm{kg}$ glider rides a $25\,\mathrm{N}/\mathrm{m}$ spring: compute $T_0$ and $f_0$. What does doubling the mass do to the [period](#def-g12-oscillators-and-time-oscillator)?

**Solution of Exercise 28.3.**

$T_0 = 2\pi\sqrt{0.10/25} = 0.40\,\mathrm{s}$, $f_0 = 2.5\,\mathrm{Hz}$. Doubling $m$: $T_0 \times \sqrt{2} \approx 1.4$.

**Exercise 28.4 ★.**

Successive maxima of a [damped](#def-g12-oscillators-and-time-damping) [oscillator](#def-g12-oscillators-and-time-oscillator): $2.0$, $1.5$, $1.13$, $0.85\,\mathrm{cm}$ at $t = 0$, $2.0$, $4.0$, $6.0\,\mathrm{s}$. Read the [pseudo-period](#def-g12-oscillators-and-time-damping); show the [amplitude](#def-g12-oscillators-and-time-oscillator) falls by a fixed factor each [period](#def-g12-oscillators-and-time-oscillator); predict the next maximum.

**Solution of Exercise 28.4.**

[Pseudo-period](#def-g12-oscillators-and-time-damping) $2.0\,\mathrm{s}$. Ratios $1.5/2.0 = 1.13/1.5 = 0.85/1.13
= 0.75$ each [period](#def-g12-oscillators-and-time-oscillator); next maximum $0.85 \times 0.75 \approx
0.64\,\mathrm{cm}$.

**Exercise 28.5 ★.**

An astronaut carries a pendulum and a [spring–mass oscillator](#prop-g12-oscillators-and-time-spring-period) to the Moon ($g = 1.62\,\mathrm{m}/\mathrm{s}^{2}$). By what factor does each [period](#def-g12-oscillators-and-time-oscillator) change? Explain.

**Solution of Exercise 28.5.**

Pendulum: $T_0 \propto 1/\sqrt{g}$, so $\times\sqrt{9.81/1.62} =
2.46$ (slower). Spring–mass: unchanged — no $g$ in $2\pi\sqrt{m/k}$.

**Exercise 28.6 ★★.**

(a) Check that $\sqrt{L/g}$ has the dimension of a time. (b) Why can no formula for $T_0$ built from $m$, $L$, $g$ contain the mass? (c) Could this argument ever produce the $2\pi$?

**Solution of Exercise 28.6.**

(a) $[L/g] = \mathrm{m}/(\mathrm{m}/\mathrm{s}^{2}) = \mathrm{s}^{2}$; its square root is a time. (b) The [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) appears only in $m$: nothing else could cancel it, so $m$ cannot enter. (c) Never: pure numbers like $2\pi$ are invisible to dimensions.

**Exercise 28.7 ★★.**

Verify by differentiating twice that $x(t) = A\cos(2\pi t/T_0 +
\varphi)$ satisfies $\frac{d^2x}{dt^2} = -(k/m)\,x$ if and only if $T_0 = 2\pi\sqrt{m/k}$.

**Solution of Exercise 28.7.**

$\frac{dx}{dt} = -\frac{2\pi}{T_0}A\sin(2\pi t/T_0 + \varphi)$, so $\frac{d^2x}{dt^2} = -(2\pi/T_0)^2 x(t)$. Equality with $-(k/m)x$ for all $t$ holds exactly when $(2\pi/T_0)^2 = k/m$, i.e. $T_0 = 2\pi\sqrt{m/k}$.

**Exercise 28.8 ★★.**

A spring of [stiffness](#def-g12-oscillators-and-time-spring) $40\,\mathrm{N}/\mathrm{m}$ carries a $0.20\,\mathrm{kg}$ glider with [amplitude](#def-g12-oscillators-and-time-oscillator) $3.0\,\mathrm{cm}$. Compute the [mechanical energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em) and the maximum speed. Where is the speed maximal? Zero?

**Solution of Exercise 28.8.**

$E_m = \tfrac12 kA^2 = \tfrac12 \times 40 \times 0.030^2 =
1.8 \times 10^{-2}\,\mathrm{J}$; $v_{\max} = \sqrt{2E_m/m} = \sqrt{0.18} =
0.42\,\mathrm{m}/\mathrm{s}$ — at $x = 0$; zero at the extremes $x = \pm A$.

**Exercise 28.9 ★★.**

(a) Compute the length of the seconds pendulum ($T_0 =
2.00\,\mathrm{s}$). (b) Its length grows by $1.0\%$: with $\sqrt{1+x}
\approx 1 + x/2$, find the relative change of $T_0$ and the daily error.

**Solution of Exercise 28.9.**

(a) $L = gT_0^2/4\pi^2 = 0.994\,\mathrm{m}$. (b) $\Delta T_0/T_0 =
\tfrac12 \times 1.0\% = 0.5\%$: $86400 \times 0.005 \approx
4.3 \times 10^{2}\,\mathrm{s}$ — seven minutes slow per day.

**Exercise 28.10 ★★.**

A lightly [damped](#def-g12-oscillators-and-time-damping) [oscillator](#def-g12-oscillators-and-time-oscillator) loses $5.0\%$ of its [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) each [period](#def-g12-oscillators-and-time-oscillator). (a) The [amplitude](#def-g12-oscillators-and-time-oscillator) drop per [period](#def-g12-oscillators-and-time-oscillator)? (b) How many [periods](#def-g12-oscillators-and-time-oscillator) to lose half the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy)? (c) Is the [pseudo-period](#def-g12-oscillators-and-time-damping) far from $T_0$?

**Solution of Exercise 28.10.**

(a) $A \propto \sqrt{E}$: $\sqrt{0.95} \approx 0.975$, a $2.5\%$ drop. (b) $0.95^n = 0.5$: $n = \ln 2/\ln(1/0.95) \approx 14$ [periods](#def-g12-oscillators-and-time-oscillator). (c) No: for light [damping](#def-g12-oscillators-and-time-damping) it stays within a fraction of a percent of $T_0$.

**Exercise 28.11 ★★.**

Explain with the [resonance](#def-g12-oscillators-and-time-resonance) curve: (a) why pushing a swing [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) only at its own rhythm; (b) why soldiers break step on footbridges; (c) why a washing machine shudders at one speed while spinning up.

**Solution of Exercise 28.11.**

(a) At $f_0$ each push arrives in step and adds [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy); off the peak, pushes alternately help and hinder. (b) Marching in step is a [periodic](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-period) drive; near the bridge’s $f_0$ it would climb the [resonance](#def-g12-oscillators-and-time-resonance) peak. (c) The drum sweeps through the frame’s [natural frequency](#def-g12-oscillators-and-time-resonance) during spin-up: momentary [resonance](#def-g12-oscillators-and-time-resonance), then past the peak.

**Exercise 28.12 ★★★.**

A pendulum clock, exact at sea level, moves to an observatory where $g = 9.78\,\mathrm{m}/\mathrm{s}^{2}$. (a) Fast or slow? By how many seconds per day? (b) How much shorter must its $0.994\,\mathrm{m}$ pendulum be?

**Solution of Exercise 28.12.**

(a) $T_0 \propto 1/\sqrt{g}$: slow, by $\Delta T_0/T_0 = \tfrac12
\times 0.03/9.81 = 1.5 \times 10^{-3}$, i.e. $\approx 132\,\mathrm{s}$ per day. (b) $L \propto g$ at fixed $T_0$: shorten by $0.994 \times
0.03/9.81 \approx 3.0\,\mathrm{mm}$.

**Exercise 28.13 ★★★.**

A $0.20\,\mathrm{kg}$ glider obeys $x(t) = A\cos(2\pi t/T_0)$, with $A =
4.0\,\mathrm{cm}$, $T_0 = 0.63\,\mathrm{s}$. Compute the maximum speed, the maximum acceleration, the [stiffness](#def-g12-oscillators-and-time-spring) $k$ and the [mechanical energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em).

**Solution of Exercise 28.13.**

$v_{\max} = 2\pi A/T_0 = 2\pi \times 0.040/0.63 = 0.40\,\mathrm{m}/\mathrm{s}$; $a_{\max} = (2\pi/T_0)^2 A = 4.0\,\mathrm{m}/\mathrm{s}^{2}$; $k = m(2\pi/T_0)^2
= 20\,\mathrm{N}/\mathrm{m}$; $E_m = \tfrac12 kA^2 = 1.6 \times 10^{-2}\,\mathrm{J}$.

**Exercise 28.14 ★★★.**

To measure $g$: a pendulum with $L = 99.5 \pm 0.2\,\mathrm{cm}$ makes $20$ oscillations in $40.1 \pm 0.2\,\mathrm{s}$. Compute $g$ and its [relative uncertainty](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-uncertainty) (add those of $L$ and $T_0^2$); does the result agree with $9.81\,\mathrm{m}/\mathrm{s}^{2}$?

**Solution of Exercise 28.14.**

$T_0 = 40.1/20 = 2.005\,\mathrm{s}$; $g = 4\pi^2 L/T_0^2 = 4\pi^2
\times 0.995/4.020 = 9.77\,\mathrm{m}/\mathrm{s}^{2}$. Uncertainty $\approx 0.2\%
+ 2 \times 0.5\% = 1.2\%$, i.e. $\pm0.12\,\mathrm{m}/\mathrm{s}^{2}$: the interval contains $9.81\,\mathrm{m}/\mathrm{s}^{2}$ — consistent.

**Exercise 28.15 ★★★.**

Compare timekeepers by relative drift (error over elapsed time): a pendulum clock losing $10\,\mathrm{s}$ per day, a quartz watch $0.5\,\mathrm{s}$ per month, a cesium clock $1\,\mathrm{s}$ per hundred million years: compute the three ratios. A GPS clock wrong by $1\,\text{µ}\mathrm{s}$: position error? Which family does GPS need?

**Solution of Exercise 28.15.**

Pendulum $10/86400 \approx 1.2 \times 10^{-4}$; quartz $0.5/2.6 \times 10^{6}
\approx 1.9 \times 10^{-7}$; cesium $1/3.2 \times 10^{15} \approx 3 \times 10^{-16}$. A $1\,\text{µ}\mathrm{s}$ error is $1000 \times 30\,\mathrm{cm} =
300\,\mathrm{m}$ of position: GPS needs the atomic family.

## 28.7 Problem: The Clockmaker’s Bench

**Problem 28.1.**

Weekend problem — the clockmaker’s bench: a wall clock brought back to time — fifty swings timed, a brass rod that stretches in summer, the escapement’s whispered push, a quartz challenger — the verdict rendered in seconds per month

A pendulum wall clock arrives at the bench “running wrong”. Treat its brass rod and heavy bob as a [simple pendulum](#def-g12-oscillators-and-time-pendulum) of effective length $L$, bob mass $m = 1.2\,\mathrm{kg}$; the gearing counts each full oscillation as exactly $2.000\,\mathrm{s}$.

**Part I — Taking the clock’s pulse.**

1. Why time $50$ oscillations rather than one? Estimate the error left on $T_0$ if your reaction time is $0.2\,\mathrm{s}$ .
2. The stopwatch reads $99.4\,\mathrm{s}$ for $50$ oscillations: the [period](#def-g12-oscillators-and-time-oscillator) ?
3. Deduce the present effective length of the pendulum.
4. What length would give exactly $2.000\,\mathrm{s}$ ? Should the rating nut raise or lower the bob, and by how much?
5. Unadjusted, is the clock fast or slow, by how many seconds per day?
6. Sensitivity to $g$ : the [period](#def-g12-oscillators-and-time-oscillator) on the Moon ( $g = 1.62\,\mathrm{m}/\mathrm{s}^{2}$ )? The daily error at an observatory where $g = 9.80\,\mathrm{m}/\mathrm{s}^{2}$ ?

**Part II — The summer slump.** Brass expands: a warming $\Delta\theta$ stretches the rod to $L' = L\,(1 + \lambda\,
\Delta\theta)$, $\lambda = 1.9 \times 10^{-5}$ per [kelvin](https://one-course.com/books/physics/2/en/chapter/2-light-spectra-and-the-message-of-light#def-g10-light-spectra-kelvin). The clock is adjusted at $20\,{}^{\circ}\mathrm{C}$.

7. Compute $\Delta L$ for a summer room at $25\,{}^{\circ}\mathrm{C}$ .
8. Using $\sqrt{1+x} \approx 1 + x/2$ , show that $\Delta T_0 / T_0 = \tfrac12 \lambda\, \Delta\theta$ .
9. Compute $\Delta T_0/T_0$ and the daily summer error: fast or slow?
10. Same questions for a winter hallway at $10\,{}^{\circ}\mathrm{C}$ .
11. Old precision clocks used “gridiron” pendulums mixing brass rods with steel ones (which expand about half as much) in alternating directions. Explain how this can hold the rate.

**Part III — The whispered push.** With the escapement disengaged, the swing, started at [amplitude](#def-g12-oscillators-and-time-oscillator) $\theta_m = 2.00^\circ$, halves its [amplitude](#def-g12-oscillators-and-time-oscillator) in $280\,\mathrm{s}$; in service, a tiny push from the escapement each [period](#def-g12-oscillators-and-time-oscillator) keeps $\theta_m$ constant.

12. How many [periods](#def-g12-oscillators-and-time-oscillator) does the halving take?
13. At the extreme the bob has climbed $h = L(1 - \cos\theta_m)  \approx L\,\theta_m^2/2$ (radians): deduce $E_m = \tfrac12  m g L\,\theta_m^2$ and compute it ( [Chapter 18](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#ch-g11-mechanical-energy) ).
14. [Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) is proportional to [amplitude](#def-g12-oscillators-and-time-oscillator) squared: what fraction of $E_m$ is lost per [period](#def-g12-oscillators-and-time-oscillator) ?
15. Deduce the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) of each push, and the average [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) needed.
16. The drive [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $M = 2.5\,\mathrm{kg}$ descends $h =  1.0\,\mathrm{m}$ per week: is that enough [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) , at what [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) ?

**Part IV — The quartz challenger.** A quartz movement oscillates at $f = 32\,768\,\mathrm{Hz}$ and drifts about $0.3\,\mathrm{s}$ per month; the restored pendulum still drifts $0.5\,\mathrm{s}$ per day.

17. Compute the quartz [period](#def-g12-oscillators-and-time-oscillator) . Show $32\,768 = 2^{15}$ : why is a [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) of two exactly what a watch circuit wants?
18. Compute the relative drifts of quartz and restored pendulum, and each in seconds per month. Who wins, by what factor?
19. The cesium standard — $9\,192\,631\,770$ [periods](#def-g12-oscillators-and-time-oscillator) make one second — drifts $1\,\mathrm{s}$ per hundred million years: its relative drift? How many orders of magnitude below quartz, and why does GPS demand it?
20. The verdict, one sentence with numbers: what does the restored clock earn on the wall, and who now owns the second?

**Solution of Problem 28.1.**

**1.** The $\approx 0.2\,\mathrm{s}$ reaction error is divided by $50$: about $4\,\mathrm{ms}$ on $T_0$.

**2.** $T = 99.4/50 = 1.988\,\mathrm{s}$.

**3.** $L = gT^2/4\pi^2 = 9.81 \times 1.988^2/4\pi^2 \approx
0.982\,\mathrm{m}$.

**4.** $L = 0.994\,\mathrm{m}$ (the seconds pendulum): lower the bob, lengthening by $\approx 1.2\,\mathrm{cm}$.

**5.** Too short, so too quick: fast, by $(2.000 -
1.988)/1.988 \times 86400 \approx 5.2 \times 10^{2}\,\mathrm{s} \approx 8.7$ minutes per day.

**6.** Moon: $\times\sqrt{9.81/1.62} = 2.46$ — unusable. Observatory: $\Delta T_0/T_0 = \tfrac12 \times 0.01/9.81 =
5.1 \times 10^{-4}$: loses $\approx 44\,\mathrm{s}$ per day.

**7.** $\Delta L = L\lambda\,\Delta\theta = 0.994 \times
1.9 \times 10^{-5} \times 5.0 \approx 9.4 \times 10^{-5}\,\mathrm{m} \approx
0.1\,\mathrm{mm}$.

**8.** $T' = 2\pi\sqrt{L(1 + \lambda\Delta\theta)/g} =
T_0\sqrt{1 + \lambda\Delta\theta} \approx T_0\,(1 + \tfrac12
\lambda\Delta\theta)$.

**9.** $\Delta T_0/T_0 = \tfrac12 \times 1.9 \times 10^{-5} \times
5.0 = 4.8 \times 10^{-5}$: $86400 \times 4.8 \times 10^{-5} \approx
4.1\,\mathrm{s}$ per day, slow (longer rod, longer [period](#def-g12-oscillators-and-time-oscillator)).

**10.** $\Delta\theta = -10\,\mathrm{K}$: $\Delta T_0/T_0 =
-9.5 \times 10^{-5}$, gains $\approx 8.2\,\mathrm{s}$ per day.

**11.** Steel and brass expand differently; hung in alternating directions, the brass rods lower the bob while the steel ones raise it. Sized so the two shifts cancel, the effective length — and the rate — survives the seasons.

**12.** $280/2.000 = 140$ [periods](#def-g12-oscillators-and-time-oscillator).

**13.** $\theta_m = 2.00^\circ = 0.0349\,\mathrm{rad}$: $E_m =
\tfrac12 \times 1.2 \times 9.81 \times 0.994 \times 0.0349^2 \approx
7.1 \times 10^{-3}\,\mathrm{J}$.

**14.** Per [period](#def-g12-oscillators-and-time-oscillator) the [amplitude](#def-g12-oscillators-and-time-oscillator) shrinks by $2^{-1/140} \approx
0.9951$, the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) by $0.9951^2 \approx 0.990$: about $1.0\%$ of $E_m$ per [period](#def-g12-oscillators-and-time-oscillator).

**15.** Push $\approx 0.010 \times 7.1 \times 10^{-3} =
7.1 \times 10^{-5}\,\mathrm{J}$; [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) $7.1 \times 10^{-5}/2.0 \approx 3.5 \times 10^{-5}\,\mathrm{W}$ — tens of microwatts.

**16.** A week is $604800/2.0 = 3.02 \times 10^{5}$ [periods](#def-g12-oscillators-and-time-oscillator), needing $3.02 \times 10^{5} \times 7.1 \times 10^{-5} \approx 21\,\mathrm{J}$; the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) supplies $Mgh = 2.5 \times 9.81 \times 1.0 = 24.5\,\mathrm{J}$: enough, at $21/24.5 \approx 88\%$ [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency).

**17.** $T = 1/32\,768 = 30.5\,\text{µ}\mathrm{s}$. $32\,768
= 2^{15}$: fifteen divide-by-two stages turn the crystal’s [signal](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal) into an exact $1\,\mathrm{Hz}$ tick.

**18.** Quartz: $0.3/2.6 \times 10^{6} \approx 1.2 \times 10^{-7}$. Pendulum: $0.5/86400 \approx 5.8 \times 10^{-6}$, i.e. $\approx
15\,\mathrm{s}$ per month. Quartz wins by a factor $\approx 50$.

**19.** $1/3.2 \times 10^{15} \approx 3 \times 10^{-16}$: nine orders of magnitude below quartz. GPS times [signals](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal) to nanoseconds — $30\,\mathrm{cm}$ of light travel each — so only atomic clocks will do.

**20.** Restored, the wall clock holds $\pm15\,\mathrm{s}$ per month against the quartz’s $\pm0.3\,\mathrm{s}$: keep it for the chime, trust the quartz for the train — and the second itself now belongs to the cesium [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder).