E. Waves and Sound · Waves and SoundThe Speed of Sound in Solids, Liquids and Gases
E3.5 E2.4 E2.1 A1.13About 65 minutes
Learning goals- I can explain, using the particle model, why sound generally travels fastest in solids and slowest in gases.
- I can calculate the speed of sound in air at any temperature with v = 331 + 0.6T.
- I can solve problems involving sound travelling different distances or through different media.
- I can use the delay between lightning and thunder to estimate distance.
Words to know
speed of sound
How fast a sound wave travels through a medium; in air about 343 m/s at 20 °C.
elasticity (stiffness)
How strongly the particles of a material push back when they are squeezed together. Stiffer materials carry sound faster.
density
Mass per unit volume. More inertia (mass) makes particles slower to respond, which by itself slows sound down.
temperature
A measure of the average kinetic energy of the particles. Warmer air carries sound faster.
Count the seconds
On a summer evening at the cottage you see a flash of lightning over the lake, then you count: one, two, three… and the thunder rolls in. Light reaches you almost instantly, but sound takes its time. How fast does it go, and why does that speed change from air to water to steel? The answers come from the particle model of matter.
Why the medium matters
Sound travels as each particle pushes on its neighbours. Two properties of the medium decide how fast this happens:
- How strongly particles are linked (elasticity or stiffness). In a solid, particles are held tightly by strong forces, so a push is passed on almost at once. In a gas, particles are far apart and only pass on a push when they happen to collide. Stiffer media carry sound faster.
- How much inertia the particles have (density). Heavier particles are harder to get moving, which by itself slows sound down.
In practice, stiffness wins: sound is usually fastest in solids, slower in liquids and slowest in gases. Among gases, ones with light molecules such as helium carry sound much faster than air, which is why helium makes voices sound squeaky: it changes the resonances of your throat and mouth (the source's vibration frequency stays about the same).
Temperature and the speed of sound in air
In a gas, the push is passed on by moving molecules colliding. Warmer molecules move faster, so they pass the disturbance along faster. Near everyday temperatures the speed of sound in dry air is
v = 331 m/s + (0.6 m/s/°C) T
where T is the air temperature in degrees Celsius. Each 1 °C increase adds about 0.6 m/s. At 0 °C, v = 331 m/s; at 20 °C, v = 343 m/s. (Some textbooks use 0.59 or 0.606 instead of 0.6; answers differ only slightly.)
Theory versus measurement
The formula gives a theoretical value. When you measure the speed of sound in an investigation, compare your result with it using percent difference = |measured − theoretical| ÷ theoretical × 100%. Then explain the difference: reaction time, a poor distance measurement, wind, humidity or temperature differences are common causes.
Worked exampleExample 1: Winter and summer
Find the speed of sound in air on a −15 °C January day in Ottawa and on a 30 °C day in July.
- Winter: v = 331 + 0.6(−15) = 331 − 9 = 322 m/s.
- Summer: v = 331 + 0.6(30) = 331 + 18 = 349 m/s.
- Sound is about 27 m/s (about 8%) faster in summer.
322 m/s in January and 349 m/s in July.
Worked exampleExample 2: How far away is the storm?
You see lightning and hear the thunder 6.0 s later. The air temperature is 20 °C. How far away was the lightning?
- Light takes only about 0.00002 s to travel a few kilometres, so treat the flash as instant.
- v = 331 + 0.6(20) = 343 m/s.
- d = vt = 343 × 6.0 = 2058 m ≈ 2.1 km.
- Quick rule: sound travels about 1 km every 3 s, so 6 s ≈ 2 km. ✓
Lightning and thunder. Divide the delay in seconds by 3 to estimate the distance in kilometres. (Safety rule from Environment and Climate Change Canada: when thunder roars, go indoors.)Worked exampleExample 3: Two sounds from one tap
A friend taps one end of a 100 m steel park railing with a stone. You have your ear near the other end (not touching). Sound travels through the steel at 5000 m/s and through the air at 343 m/s. What is the time between the two sounds you hear?
- Time through steel: t = d/v = 100 ÷ 5000 = 0.020 s.
- Time through air: t = 100 ÷ 343 = 0.2915 s.
- Difference: 0.2915 − 0.020 = 0.27 s.
About 0.27 s: you hear a faint tap through the railing first, then the louder sound through the air.
Worked exampleExample 4: Find the temperature
An investigation finds the speed of sound in a gym to be 340 m/s. What temperature does the formula predict?
- Rearrange: T = (v − 331) ÷ 0.6.
- T = (340 − 331) ÷ 0.6 = 9 ÷ 0.6 = 15 °C.
Just for funFunny-but-true: on a cold winter day, sound is a little slower than in summer, about 320 m/s at −20 °C. So your "Hurry up, we're late!" really does take longer to reach someone in January.
Hands-on activityMeasure the speed of sound with echoes
You need: a large flat wall with open space in front of it (a school wall on a weekend works well), a measuring tape or a phone map app to measure about 50 m to 100 m, a stopwatch, an outdoor thermometer (or a weather app), two wooden blocks or your hands, and a partner.
- Stand at least 50 m from the wall. Measure the distance d.
- Clap the blocks together. Listen for the echo.
- Now clap steadily so that each clap happens at the same moment the echo of the previous clap returns. This takes some practice.
- When the rhythm is steady, your partner times 20 claps (start counting at zero).
- Between claps, sound goes to the wall and back: t = (total time) ÷ 20. Then v = 2d ÷ t.
- Record the air temperature. Calculate the theoretical speed with v = 331 + 0.6T and the percent difference.
What to look for: with d = 60 m, each clap interval is about 0.35 s, so 20 claps take about 7 s. A good result is within about 10% of the theoretical value. Discuss your largest sources of error (reaction time, wind, measuring d).
Your turn
1Rank sound speed in water, steel and air from fastest to slowest.
2Find the speed of sound in air at 10 °C.
v = 331 + 0.6(10) = 337 m/s.
3Find the speed of sound in air at −25 °C.
v = 331 + 0.6(−25) = 331 − 15 = 316 m/s.
4Explain why sound travels faster in warm air than in cold air.
Warmer air molecules move faster, so they collide and pass the pressure disturbance along more quickly.
5Thunder is heard 9.0 s after a lightning flash at 25 °C. How far away was the strike?
v = 331 + 0.6(25) = 346 m/s. d = 346 × 9.0 = 3114 m ≈ 3.1 km.
6Why is sound generally faster in solids than in gases, even though solids are denser?
Particles in solids are linked by much stronger forces (higher stiffness), so a push passes on very quickly. This effect is far bigger than the slowing effect of higher density.
7At what temperature is the speed of sound 349 m/s?
T = (349 − 331) ÷ 0.6 = 30 °C.
More practice: warm-up
1What is the speed of sound in air at 0 °C?
2By about how much does the speed of sound in air change for each 1 °C?
3In which state of matter does sound usually travel slowest?
4Sound takes 3 s to travel about how far in air?
About 1 km (343 × 3 ≈ 1029 m).
5Can sound travel through a vacuum?
No; it needs particles to pass the vibration along.
More practice: on your own
1A student measures the speed of sound as 352 m/s on a 22 °C day. Find the theoretical value and the percent difference.
Theoretical: 331 + 0.6(22) = 344.2 m/s. Percent difference = |352 − 344.2| ÷ 344.2 × 100% ≈ 2.3%.
2How long does sound take to cross a 1.5 km wide lake at 5 °C?
v = 331 + 0.6(5) = 334 m/s. t = 1500 ÷ 334 ≈ 4.5 s.
3Sound travels 1480 m/s in a lake. How long does a sound take to travel 2.0 km through the water? Compare with air at 20 °C.
Water: 2000 ÷ 1480 ≈ 1.35 s. Air: 2000 ÷ 343 ≈ 5.83 s. Water is about 4.3 times faster.
4In the clap-echo method, a student stands 85 m from a wall and times 20 clap intervals as 10.0 s. Find the speed of sound.
t per round trip = 10.0 ÷ 20 = 0.500 s. v = 2(85) ÷ 0.500 = 340 m/s.
5A 512 Hz sound passes from air at 20 °C into steel (5900 m/s). Find its wavelength in each.
Air: 343 ÷ 512 ≈ 0.670 m. Steel: 5900 ÷ 512 ≈ 11.5 m. Same frequency, much longer wavelength in steel.
Challenge
1At a large outdoor concert, a fan sits 170 m from the stage. The air is 20 °C. Another fan listens to the same concert on the radio 3000 km away (radio waves travel at 3.00 × 10⁸ m/s, and assume the radio signal is sent the instant the microphone picks up the sound). Who hears a note first, and by how much?
Fan at concert: t = 170 ÷ 343 ≈ 0.496 s. Radio fan: t = 3.0 × 10⁶ ÷ 3.00 × 10⁸ = 0.010 s (plus a tiny time for the microphone's distance, ignored). The radio listener hears it first, by about 0.49 s.
2A hiker shouts toward a cliff and hears the echo 2.40 s later at −10 °C. On a 25 °C day the same echo takes how long?
At −10 °C: v = 331 − 6 = 325 m/s; distance to cliff = 325 × 2.40 ÷ 2 = 390 m. At 25 °C: v = 346 m/s; t = 2(390) ÷ 346 ≈ 2.25 s.
3A student hits one end of a long metal pipe. A friend at the other end hears two sounds 0.75 s apart. Sound travels at 343 m/s in air and 5000 m/s in the pipe. How long is the pipe?
L/343 − L/5000 = 0.75 → L(0.0029155 − 0.0002) = 0.75 → L(0.0027155) = 0.75 → L ≈ 276 m (about 2.8 × 10² m).
Quiz
1. In general, sound travels fastest in…
2. The speed of sound in air at 15 °C is…
3. Why is sound faster in warm air?
4. Thunder arrives 12 s after the flash. The storm is about…
5. A sound passes from air into water. Which statement is correct?
6. A measured speed of sound is 330 m/s; the theoretical value is 344 m/s. The percent difference is about…
Big idea: expectation E3.5 asks students to explain how the speed of sound depends on the medium (solid, liquid, gas) and on temperature, using the particle nature of matter. E2.3 (measuring the speed of sound and comparing theory with measurement) is introduced here and done fully in the air-column lesson.
How to help: Count lightning-to-thunder seconds together during a storm (from indoors!). The clap-echo activity is a great family investigation.
Common difficulties:
- Thinking denser always means faster. Stiffness matters more.
- Using Kelvin or Fahrenheit in v = 331 + 0.6T (it needs °C).
- Forgetting to double the distance in echo methods.
- Mixing up which time is longer in two-medium problems: the slower medium takes longer.