High School Physics · Grades 10–12

8Signals and Waves

Every message you have ever received arrived as a signal: a voice pressed on your eardrum, this page reached you as a flicker of radio. Behind the variety sits one idea — a physical quantity varying in time — and one toolkit: period, frequency, amplitude, and the art of timing a delay. With it, a boat measures the sea floor and a doctor counts heartbeats.

8.1 Signals carry information

Definition 8.1 (Signal)

A signal is a physical quantity that varies in time to carry information: the air pressure at your eardrum, the voltage at a microphone’s terminals, the brightness of the light in an optical fiber. To study a signal, record it as a function of time and read the information off the curve.

In a phone call the same information changes carrier repeatedly: pressure in air, then voltage in the microphone, then radio at the antenna — and backwards at the other end. One vocabulary handles them all; this chapter builds it.

8.2 Periodic signals

Definition 8.2 (Periodic signal, period)

A signal is periodic when it repeats identically at regular intervals. The period $T$ is the duration of one complete cycle, in seconds.

Definition 8.3 (Frequency)

The frequency of a periodic signal is the number of cycles per second:

f = \frac{1}{T} .

Its unit, one cycle per second, is the hertz ( $\mathrm{Hz}$ ), with the usual multiples $\mathrm{kHz}$ , $\mathrm{MHz}$ , $\mathrm{GHz}$ .

Definition 8.4 (Amplitude)

The amplitude of a periodic signal is its maximal deviation from the resting value. For a voltage oscillating symmetrically about zero, the amplitude is the peak value; crest to trough measures twice the amplitude (the peak-to-peak value).

Example 8.5 (Mains and concert pitch)

The mains voltage oscillates at $f = 50\,\mathrm{Hz}$ : its period is $T = 1/f = 0.020\,\mathrm{s} = 20\,\mathrm{ms}$ . An orchestra’s concert A is a pressure signal at $440\,\mathrm{Hz}$ : one cycle lasts $T = 1/440 = 2.3\,\mathrm{ms}$ . Higher frequency, shorter cycle: $f$ and $T$ are inverses of each other.

A periodic signal: the pattern repeats every T seconds, and the amplitude is the maximal deviation from the resting value. — A periodic signal: the pattern repeats every $T$ seconds, and the amplitude is the maximal deviation from the resting value.

8.3 Reading an oscillogram

Definition 8.6 (Oscilloscope)

An oscilloscope plots a voltage against time; the curve on its gridded screen is an oscillogram. Two settings convert screen divisions into physical values: the time base (seconds per horizontal division) and the vertical gain (volts per vertical division).

Method 8.7 (From screen to numbers)

Count the horizontal divisions spanned by one complete cycle (crest to crest is safest); multiply by the time base to get $T$ , then $f = 1/T$ .
Count the vertical divisions from the center line to a crest and multiply by the gain to get the amplitude (crest to trough gives the peak-to-peak value: halve it).

Example 8.8 (A full reading)

On the oscillogram below, the time base is $5\,\mathrm{ms}$ per division and the gain $2\,\mathrm{V}$ per division. One cycle spans $4.0$ divisions: $T = 4.0 \times 5\,\mathrm{ms} = 20\,\mathrm{ms}$ and $f = 1/0.020\,\mathrm{s} = 50\,\mathrm{Hz}$ — the mains again. A crest sits $3.0$ divisions above the axis: the amplitude is $3.0 \times 2\,\mathrm{V} = 6.0\,\mathrm{V}$ .

An oscillogram, time base 5\, ms/div and gain 2\, V/div: one cycle spans 4.0 divisions (T = 20\, ms, f = 50\, Hz) and the crest sits 3.0 divisions up (amplitude 6.0\, V). — An oscillogram, time base $5\,\mathrm{ms}$ /div and gain $2\,\mathrm{V}$ /div: one cycle spans $4.0$ divisions ( $T = 20\,\mathrm{ms}$ , $f = 50\,\mathrm{Hz}$ ) and the crest sits $3.0$ divisions up (amplitude $6.0\,\mathrm{V}$ ).

8.4 Sound and electromagnetic signals

Definition 8.9 (Sound)

Sound is a pressure signal: a vibrating object — string, membrane, vocal cords — pushes rhythmically on the air, and the compressions travel outward at about $340\,\mathrm{m}/\mathrm{s}$ . A microphone converts the arriving pressure into a voltage; the eardrum, into nerve impulses. Sound needs a medium: a bell rung under a vacuum jar falls silent as the air is pumped out.

Definition 8.10 (Wave)

A traveling perturbation — a sound’s compression pattern, the ripple on a pond, a radio pulse — is a wave. The medium, when there is one, stays put: each patch of water bobs in place while the ripple crosses the pond. The geometry of waves — wavelength, interference, diffraction — is taken up in a later chapter; this chapter only needs their speeds.

Definition 8.11 (Ultrasound and infrasound)

The human ear hears sound between about $20\,\mathrm{Hz}$ and $20\,\mathrm{kHz}$ (the ceiling drops with age). Sound above $20\,\mathrm{kHz}$ is ultrasound — bats hunt at $50\,\mathrm{kHz}$ , medical scanners image the body at a few megahertz. Sound below $20\,\mathrm{Hz}$ is infrasound, felt by elephants and seismometers.

Proposition 8.12 (Electromagnetic signals)

Radio, Wi-Fi and light are electromagnetic signals: one family, all traveling through vacuum — no medium needed — and all at one speed, the speed of light of Chapter 3,

c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s} \quad \text{(very nearly the same in air).}

Proof. Admitted at this level. ∎

Remark 8.13

Here this is an experimental fact — sunlight crosses $150$ million kilometres of empty space; it is honestly derived in the Year 2 volume from the laws of electricity and magnetism. Keep the contrast: sound is a vibration of the medium and dies without it; an electromagnetic signal carries its own field across vacuum, a million times faster.

Sound frequencies on a logarithmic scale (each step × 10). The ear hears from 20\, Hz to 20\, kHz; below is infrasound, above is ultrasound. — Sound frequencies on a logarithmic scale (each step $\times 10$ ). The ear hears from $20\,\mathrm{Hz}$ to $20\,\mathrm{kHz}$ ; below is infrasound, above is ultrasound.

8.5 Echoes: distances from delays

A signal moving at speed $v$ covers $d = v\,t$ in a time $t$ : every delay is a distance in disguise. Better still, send a short pulse at an obstacle and time its echo — the round trip covers $2d$ .

Method 8.14 (Echo ranging)

Send a short pulse; measure the round-trip time $t$ of the echo.
Take the speed $v$ of the signal in the medium crossed.
The pulse covered $2d = v\,t$ , so $d = \dfrac{v\,t}{2}$ .

Sound pulses in water make sonar ( $v \approx 1500\,\mathrm{m}/\mathrm{s}$ ); radio pulses in air make radar ( $v = c$ ); ultrasound pulses in the body make the medical scanner ( $v \approx 1540\,\mathrm{m}/\mathrm{s}$ in soft tissue).

Example 8.15 (Sonar)

A ship’s sonar pings and the seabed echo returns after $t = 0.80\,\mathrm{s}$ : the depth is $d = \frac{1500 \times 0.80}{2} = 600\,\mathrm{m}$ . Forgetting the factor $2$ doubles the ocean.

A sonar ping travels down and back: the measured delay t covers 2d, so the depth is d = v\,t/2 with v 1500\, m/ s in seawater. — A sonar ping travels down and back: the measured delay $t$ covers $2d$ , so the depth is $d = v\,t/2$ with $v \approx 1500\,\mathrm{m}/\mathrm{s}$ in seawater.

Example 8.16 (Radar)

An airport radar receives an aircraft’s echo $0.30\,\mathrm{ms}$ after the pulse: $d = \frac{3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s} \times 3.0 \times 10^{-4}\,\mathrm{s}}{2} = 45\,\mathrm{km}$ . Because $c$ is so large, radar lives on microsecond clocks: timing light is precision work.

Example 8.17 (The heart as a signal)

Each heartbeat sends a small voltage pulse across the chest; electrodes record it as the electrocardiogram (ECG), a nearly periodic signal. If the tall spikes are $T = 0.75\,\mathrm{s}$ apart, the heart beats at $f = 1/0.75 = 1.33\,\mathrm{Hz}$ , i.e. $1.33 \times 60 = 80$ beats per minute: reading $T$ off a trace is how a cardiologist takes your pulse.

8.6 Exercises

Exercise 8.1 ★

Compute the frequency of the periodic signal of period (a) $T = 20\,\mathrm{ms}$ ; (b) $T = 4.0\,\mathrm{ms}$ ; (c) $T = 50\,\text{µ}\mathrm{s}$ .

Solution

Solution of Exercise 8.1.

$f = 1/T$ : (a) $1/0.020\,\mathrm{s} = 50\,\mathrm{Hz}$ ; (b) $1/4.0 \times 10^{-3}\,\mathrm{s} = 250\,\mathrm{Hz}$ ; (c) $1/5.0 \times 10^{-5}\,\mathrm{s} = 20\,\mathrm{kHz}$ .

Exercise 8.2 ★

Compute the period of (a) the $50\,\mathrm{Hz}$ mains; (b) a $440\,\mathrm{Hz}$ concert A; (c) a $2.4\,\mathrm{GHz}$ Wi-Fi signal.

Solution

Solution of Exercise 8.2.

$T = 1/f$ : (a) $20\,\mathrm{ms}$ ; (b) $1/440 = 2.3\,\mathrm{ms}$ ; (c) $1/2.4 \times 10^{9}\,\mathrm{Hz} = 0.42\,\mathrm{ns}$ .

Exercise 8.3 ★

Classify as infrasound, audible sound or ultrasound: $12\,\mathrm{Hz}$ ; $440\,\mathrm{Hz}$ ; $18\,\mathrm{kHz}$ ; $40\,\mathrm{kHz}$ (a car parking sensor); $5\,\mathrm{MHz}$ (a medical probe).

Solution

Solution of Exercise 8.3.

$12\,\mathrm{Hz}$ : infrasound. $440\,\mathrm{Hz}$ and $18\,\mathrm{kHz}$ : audible (the latter only to young ears). $40\,\mathrm{kHz}$ and $5\,\mathrm{MHz}$ : ultrasound.

Exercise 8.4 ★

On an oscilloscope set to $2\,\mathrm{ms}$ per division and $0.5\,\mathrm{V}$ per division, one cycle spans $5.0$ divisions and a crest sits $4.0$ divisions above the axis. Find $T$ , $f$ and the amplitude.

Solution

Solution of Exercise 8.4.

$T = 5.0 \times 2\,\mathrm{ms} = 10\,\mathrm{ms}$ , so $f = 100\,\mathrm{Hz}$ ; amplitude $4.0 \times 0.5\,\mathrm{V} = 2.0\,\mathrm{V}$ .

Exercise 8.5 ★

You see the lightning, then hear the thunder $6.0\,\mathrm{s}$ later. How far away did it strike ( $v_{\text{sound}} = 340\,\mathrm{m}/\mathrm{s}$ )? Why may the light’s travel time be neglected?

Solution

Solution of Exercise 8.5.

$d = 340 \times 6.0 = 2040\,\mathrm{m} \approx 2.0\,\mathrm{km}$ . Light covers this in $2040/(3.00\times10^{8}) \approx 7\,\text{µ}\mathrm{s}$ , a million times less than $6.0\,\mathrm{s}$ : the flash marks the instant of the strike.

Exercise 8.6 ★★

A sonar echo returns from the seabed after $0.60\,\mathrm{s}$ ( $v = 1500\,\mathrm{m}/\mathrm{s}$ in seawater). Why must the product $v\,t$ be halved? Compute the depth. How long would the echo take over a $1200\,\mathrm{m}$ deep trench?

Solution

Solution of Exercise 8.6.

The sound goes down and back, covering $2d$ : $d = \frac{1500 \times 0.60}{2} = 450\,\mathrm{m}$ . Over the trench: $t = 2d/v = 2 \times 1200/1500 = 1.6\,\mathrm{s}$ .

Exercise 8.7 ★★

An airport radar receives an aircraft echo after $0.24\,\mathrm{ms}$ and, exactly $1.0\,\mathrm{s}$ later, a second echo after $0.238\,\mathrm{ms}$ . Compute the two distances, then the aircraft’s speed. Is it approaching?

Solution

Solution of Exercise 8.7.

$d_1 = \frac{3.00\times10^{8} \times 2.4\times10^{-4}}{2} = 36.0\,\mathrm{km}$ and $d_2 = 35.7\,\mathrm{km}$ : the aircraft covered $300\,\mathrm{m}$ in $1.0\,\mathrm{s}$ , so $v = 300\,\mathrm{m}/\mathrm{s} \approx 1100\,\mathrm{km}/\mathrm{h}$ , approaching.

Exercise 8.8 ★★

An ECG is printed at $25\,\mathrm{mm}/\mathrm{s}$ ; the tall spikes are $20\,\mathrm{mm}$ apart. Find the period and the heart rate in beats per minute. Tachycardia means over $100$ beats per minute: below what spike spacing does it show on this paper?

Solution

Solution of Exercise 8.8.

$T = 20\,\mathrm{mm}/25\,\mathrm{mm}/\mathrm{s} = 0.80\,\mathrm{s}$ , so $f = 1.25\,\mathrm{Hz}$ : $75$ beats per minute. For $100$ beats per minute, $T = 0.60\,\mathrm{s}$ , i.e. $25 \times 0.60 = 15\,\mathrm{mm}$ : tachycardia shows below $15\,\mathrm{mm}$ .

Exercise 8.9 ★★

You must display a $200\,\mathrm{Hz}$ signal of amplitude $3.0\,\mathrm{V}$ on a screen of $10$ horizontal and $8$ vertical divisions ( $4$ above the center line). Choose a time base showing about two full cycles, and a gain using most of the screen without clipping.

Solution

Solution of Exercise 8.9.

$T = 1/200 = 5.0\,\mathrm{ms}$ ; two cycles last $10\,\mathrm{ms}$ , spread over $10$ divisions: time base $1\,\mathrm{ms}$ per division. The crest must fit in $4$ divisions: gain at least $3.0/4 = 0.75\,\mathrm{V}$ per division — take $1\,\mathrm{V}$ per division (crest $3.0$ divisions up).

Exercise 8.10 ★★

A medical probe sends an ultrasound pulse into the abdomen ( $v = 1540\,\mathrm{m}/\mathrm{s}$ in soft tissue) and hears echoes after $40\,\text{µ}\mathrm{s}$ (front wall of an organ) and $60\,\text{µ}\mathrm{s}$ (back wall). Find the depth of each wall and the organ’s thickness.

Solution

Solution of Exercise 8.10.

$d = v\,t/2$ : front wall $\frac{1540 \times 4.0\times10^{-5}}{2} = 3.1\,\mathrm{cm}$ ; back wall $\frac{1540 \times 6.0\times10^{-5}}{2} = 4.6\,\mathrm{cm}$ ; thickness about $1.5\,\mathrm{cm}$ .

Exercise 8.11 ★★

A concert is broadcast live. Who hears a drumbeat first: a listener $30\,\mathrm{m}$ from the stage, or a radio listener $3000\,\mathrm{km}$ away? By how much?

Solution

Solution of Exercise 8.11.

Sound: $30/340 = 0.088\,\mathrm{s}$ . Radio: $3.0\times10^{6}/(3.00\times10^{8}) = 0.010\,\mathrm{s}$ . The distant radio listener hears the drum first, by about $78\,\mathrm{ms}$ .

Exercise 8.12 ★★★

A bat emits ultrasound pulses lasting $3.0\,\mathrm{ms}$ and cannot hear an echo while still emitting. Closer than what distance is a moth undetectable? What is the echo delay for a moth $1.7\,\mathrm{m}$ away? Why must the bat shorten its pulses during the final approach?

Solution

Solution of Exercise 8.12.

While emitting for $3.0\,\mathrm{ms}$ the bat is deaf to echoes, i.e. to anything within $d = \frac{340 \times 3.0\times10^{-3}}{2} = 0.51\,\mathrm{m}$ . Moth at $1.7\,\mathrm{m}$ : $t = 2 \times 1.7/340 = 10\,\mathrm{ms}$ . Closing in, the echo returns ever sooner; the pulse must end before it arrives, so it must shorten.

Exercise 8.13 ★★★

You clap facing a cliff and hear the echo $1.5\,\mathrm{s}$ later. How far is the cliff? You walk $100\,\mathrm{m}$ straight toward it: new delay? The ear no longer separates clap from echo below about $0.1\,\mathrm{s}$ : within what distance does the echo disappear into the clap?

Solution

Solution of Exercise 8.13.

$d = \frac{340 \times 1.5}{2} = 255\,\mathrm{m}$ . At $155\,\mathrm{m}$ : $t = 2 \times 155/340 = 0.91\,\mathrm{s}$ . Echo and clap merge when $t < 0.1\,\mathrm{s}$ , i.e. within $d < \frac{340 \times 0.1}{2} = 17\,\mathrm{m}$ .

Exercise 8.14 ★★★

At $5\,\mathrm{ms}$ per division, one cycle of a signal spans $3.0$ divisions of a $10$ -division screen.

Find $T$ and $f$ .
The time base is switched to $2\,\mathrm{ms}$ per division: how many divisions does one cycle now span?
How many complete cycles fit on the screen at each setting?

Solution

Solution of Exercise 8.14.

1. $T = 3.0 \times 5\,\mathrm{ms} = 15\,\mathrm{ms}$ , $f = 1/0.015 \approx 67\,\mathrm{Hz}$ . 2. $15\,\mathrm{ms}/2\,\mathrm{ms} = 7.5$ divisions. 3. The screen shows $50\,\mathrm{ms}$ , then $20\,\mathrm{ms}$ : $3$ complete cycles ( $50/15 = 3.3$ ), then $1$ ( $20/15 = 1.3$ ).

Exercise 8.15 ★★★

A “ping” measures the round-trip time of an internet packet to a server $1200\,\mathrm{km}$ away. What is the smallest conceivable round-trip time (signals at $c$ )? In optical fiber, light travels at about $2.0 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ : recompute. The measured ping is $40\,\mathrm{ms}$ : give two reasons it exceeds your answers.

Solution

Solution of Exercise 8.15.

1. $t = 2 \times 1.2\times10^{6}/(3.00\times10^{8}) = 8.0\,\mathrm{ms}$ . 2. $2 \times 1.2\times10^{6}/(2.0\times10^{8}) = 12\,\mathrm{ms}$ . 3. The fiber does not run straight between the two machines, and every router and server on the way adds processing delay.

8.7 Problem: The sonar, the storm and the cardiogram

Problem 8.1

Weekend problem — reading the world’s signals: one law, $d = v\,t$ , ranges a storm, maps a seabed, counts a heartbeat and teases out how GPS works

A fishing boat works through a stormy night: lightning on the horizon, the sonar sweeping the bottom, the skipper’s heart on the doctor’s paper strip, a GPS receiver listening to satellites. Four instruments, one law. Take $v_{\text{sound}} = 340\,\mathrm{m}/\mathrm{s}$ in air, $1500\,\mathrm{m}/\mathrm{s}$ in seawater, and $c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ .

Part I — The storm.

A flash; thunder follows $9.0\,\mathrm{s}$ later. How far away is it?
Compute the light’s travel time over that distance, and justify treating the flash as instantaneous.
Sailors count the seconds between flash and thunder and divide by three to get kilometres. Justify the rule.
Five minutes later, a flash gives $6.0\,\mathrm{s}$ . How far now? Find the storm’s average approach speed in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .
Thunder rumbles instead of cracking: the lightning channel is kilometres long, so its parts lie at different distances. If the channel stretches from $2.0\,\mathrm{km}$ to $5.0\,\mathrm{km}$ from the boat, how long does the rumble last?

Part II — The sonar.

The sonar pings; the seabed answers in $0.90\,\mathrm{s}$ . Depth?
Over the shelf the echo shortens to $0.20\,\mathrm{s}$ . Depth?
One ping returns two echoes, at $0.16\,\mathrm{s}$ and $0.20\,\mathrm{s}$ . Interpret them; how far above the bottom does the fish shoal swim?
The sonar pings every $0.50\,\mathrm{s}$ . What is the greatest depth it can measure without confusion, and what goes wrong beyond it?
In air, the same $0.60\,\mathrm{s}$ delay would mean what distance? Moral: what must you know before turning a delay into a distance?

Part III — The cardiogram.

The skipper’s ECG spikes are $0.86\,\mathrm{s}$ apart. Frequency? Beats per minute?
The strip advances at $25\,\mathrm{mm}/\mathrm{s}$ . What spike spacing did the doctor measure?
After hauling nets, the spacing is $15\,\mathrm{mm}$ . New heart rate?
The young deckhand, a trained rower, rests at $50$ beats per minute. Period? Spacing on the strip?
Is an ECG strictly periodic? What, then, does the doctor read from the trace?

Part IV — Homeward by light.

The boat radios the harbor, $50\,\mathrm{km}$ away. How long does the message take? How long would sound take?
A GPS satellite orbits about $20\,200\,\mathrm{km}$ overhead. How long does its signal take to reach the boat?
GPS turns time into position. What distance error does a clock error of $1.0\,\text{µ}\mathrm{s}$ cause? What timing precision does a $3.0\,\mathrm{m}$ fix require?
Why could no sound-based GPS exist, even in principle?
Finale: list the three speeds used tonight and the single law behind all four instruments, then state the ranging habit in one sentence.

Solution

Solution of Problem 8.1.

1. $d = 340 \times 9.0 = 3060\,\mathrm{m} \approx 3.1\,\mathrm{km}$ .

2. $3060/(3.00\times10^{8}) \approx 10\,\text{µ}\mathrm{s}$ , a million times shorter than $9.0\,\mathrm{s}$ : the flash marks the instant of the strike.

3. In $3\,\mathrm{s}$ sound covers $340 \times 3 = 1020\,\mathrm{m} \approx 1\,\mathrm{km}$ : seconds divided by three gives kilometres.

4. $d = 340 \times 6.0 = 2040\,\mathrm{m}$ . The storm closed $1020\,\mathrm{m}$ in $300\,\mathrm{s}$ : $v = 3.4\,\mathrm{m}/\mathrm{s} \approx 12\,\mathrm{km}/\mathrm{h}$ , heading for the boat.

5. The near end is heard after $2000/340 = 5.9\,\mathrm{s}$ , the far end after $5000/340 = 14.7\,\mathrm{s}$ : the rumble lasts about $8.8\,\mathrm{s}$ .

6. $d = 1500 \times 0.90/2 = 675\,\mathrm{m}$ .

7. $d = 1500 \times 0.20/2 = 150\,\mathrm{m}$ .

8. Two obstacles: a fish shoal at $1500 \times 0.16/2 = 120\,\mathrm{m}$ and the seabed at $150\,\mathrm{m}$ . The shoal swims $30\,\mathrm{m}$ above the bottom.

9. The echo must return before the next ping: $d_{\max} = 1500 \times 0.50/2 = 375\,\mathrm{m}$ . From deeper water the echo arrives after the next ping and is attributed to it: the display shows a false, far too shallow bottom.

10. In air, $340 \times 0.60/2 = 102\,\mathrm{m}$ instead of $450\,\mathrm{m}$ . A delay becomes a distance only once you know the carrier and its speed in the medium crossed.

11. $f = 1/0.86 = 1.16\,\mathrm{Hz}$ , i.e. $1.16 \times 60 \approx 70$ beats per minute.

12. $25 \times 0.86 = 21.5\,\mathrm{mm}$ .

13. $T = 15/25 = 0.60\,\mathrm{s}$ : $100$ beats per minute.

14. $T = 60/50 = 1.2\,\mathrm{s}$ ; spacing $25 \times 1.2 = 30\,\mathrm{mm}$ .

15. No: the period drifts from beat to beat with effort, breathing and stress. The doctor reads the period (the rate), its regularity, and the shape of each cycle — the diagnosis is in the signal.

16. Radio: $5.0\times10^{4}/(3.00\times10^{8}) = 0.17\,\mathrm{ms}$ — instantaneous to human senses. Sound: $50\,000/340 \approx 147\,\mathrm{s}$ , two and a half minutes.

17. $t = 2.02\times10^{7}/(3.00\times10^{8}) = 67\,\mathrm{ms}$ .

18. $c \times 1.0\,\text{µ}\mathrm{s} = 300\,\mathrm{m}$ . For a $3.0\,\mathrm{m}$ fix: $3.0/(3.00\times10^{8}) = 10\,\mathrm{ns}$ — which is why GPS satellites carry atomic clocks.

19. Satellites sit in vacuum, and sound needs a medium: no signal would leave the satellite at all. (Even granting air the whole way, $20\,200\,\mathrm{km}$ at $340\,\mathrm{m}/\mathrm{s}$ is over $16\,\mathrm{h}$ — a position fix hours out of date.)

20. $340\,\mathrm{m}/\mathrm{s}$ (sound in air), $1500\,\mathrm{m}/\mathrm{s}$ (sound in seawater) and $c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ (radio and light); the law is $d = v\,t$ , halved for an echo. The habit: know your carrier, know its speed in the medium, time the delay — a delay is a distance in disguise.