PhysicsISC Class 11 · Waves

Progressive Waves: Wave Equation, Phase and the Speed of Sound

Updated 17 October 2026 · Fact-checked

A progressive wave is y = A sin(ωt − kx) with k = 2π/λ and speed v = ω/k = fλ. The speed on a string is √(T/μ), and the speed of sound in a gas is Laplace's v = √(γp/ρ), larger than Newton's √(p/ρ) by the factor √γ.

Amplitude A is the height, λ is crest to crest, and v = fλ; Laplace's adiabatic speed beats Newton's by the factor √γ.

Illustration drawn for QuestionBuddy.

Understand Progressive Waves and Speed of Sound

Types of wave. In a transverse wave the particles of the medium vibrate perpendicular to the direction of travel (waves on a string); in a longitudinal wave they vibrate along it (sound in air, forming compressions and rarefactions). A wave transfers energy and momentum without transferring matter.

Progressive wave equation. For a wave travelling along +x, y = A sin(ωt − kx) = A sin 2π(t/T − x/λ), where ω = 2πf, k = 2π/λ is the wave number, and v = ω/k = fλ. For a wave in the −x direction, the sign of kx changes. Comparing a given equation with this form gives A, f, λ and v.

Phase difference. Two points separated by a path difference Δx have a phase difference Δφ = (2π/λ)Δx. A phase difference of 2π corresponds to a distance λ.

Speed on a string. v = √(T/μ), where T is the tension and μ the mass per unit length. It increases with tension and decreases with a heavier string.

Speed of sound. Newton assumed isothermal compression and got v = √(p/ρ) ≈ 280 m s⁻¹ for air at 0 °C, lower than the measured value of about 331 m s⁻¹. Laplace pointed out that compressions are so rapid that no heat flows, so they are adiabatic: v = √(γp/ρ). In an ideal gas v ∝ √T, independent of pressure, and increases slightly with humidity and with temperature. Sound travels faster in liquids and solids.

Applications. Echo ranging uses distance = v t/2. Ultrasonic waves have frequencies above the audible range (above 20 kHz), and supersonic speeds are greater than the speed of sound.

Progressive Waves and Speed of Sound: formulas and key facts

Wave relation: $v = f λ ∣ k = 2 π / λ ∣ ω = 2 π f ∣ v = ω / k$; Progressive wave: y = A sin(ωt − kx).
Phase difference: $Δ φ = (2 π / λ) Δ x$; Δx = path difference.
String: $v = \sqrt (T / μ)$; μ = mass per unit length.
Sound in a gas: $N e w t o n : v = \sqrt (p / ρ) ∣ L a pl a ce : v = \sqrt (γ p / ρ)$; v ∝ √T.

Progressive Waves and Speed of Sound in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding the distance of a cliff

Suppose a hiker claps and hears the echo from a cliff 3.0 s later on a day when the temperature is 27 °C. The speed of sound at 0 °C (273 K) is 331 m s⁻¹.

Find the speed of sound at 27 °C and the distance to the cliff.

Try it first, then show the solutionSolution

v ∝ √T: v = 331 × √(300/273) = 331 × 1.048 = 347 m s⁻¹.
The sound travels to the cliff and back: 2d = vt.
d = 347 × 3.0/2 = 520 m.

Answer: v = 347 m s⁻¹; the cliff is about 520 m away.

Memory hook: The echo covers the distance twice.

Tricks to remember Progressive Waves and Speed of Sound

Short memory aids. Say each one aloud once, then use it on a question the same day.

k and ω from the equation
Read ω as the coefficient of t and k as the coefficient of x; v = ω/k.
Use it when: Wave equation questions.
Echo halves
Distance = speed × time / 2.
Use it when: Echo and sonar problems.

Progressive Waves and Speed of Sound: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Compressions and rarefactions · 1 mark · MCQBoard level

A wave that consists of compressions and rarefactions travelling through a medium is:
(a) transverse
(b) longitudinal
(c) electromagnetic
(d) a surface wave only

Try it first, then show the solutionSolution

The particles vibrate along the direction of travel, forming compressions and rarefactions.

Answer: (b) longitudinal

Example 2: Reading a wave equation · 3 marks · Short answerBoard level

A progressive wave is y = 0.02 sin(500t − 2x) (SI units). Find the amplitude, frequency, wavelength and speed.

Try it first, then show the solutionSolution

Compare with y = A sin(ωt − kx): A = 0.02 m, ω = 500 rad s⁻¹, k = 2 m⁻¹. (1 mark)
f = ω/2π = 500/6.283 = 79.6 Hz; λ = 2π/k = 3.14 m. (1 mark)
v = ω/k = 250 m s⁻¹; check fλ = 79.6 × 3.14 = 250 m s⁻¹. (1 mark)

Answer: A = 0.02 m; f = 79.6 Hz; λ = 3.14 m; v = 250 m s⁻¹.

Example 3: Newton and Laplace · 4 marks · Short answerHarder board level

Air at 0 °C has pressure 1.01 × 10⁵ Pa, density 1.29 kg m⁻³ and γ = 1.4. Find the speed of sound by Newton's formula and by Laplace's, and the ratio.

Try it first, then show the solutionSolution

Newton: v = √(p/ρ) = √(1.01 × 10⁵/1.29) = √78 295 = 280 m s⁻¹. (1 mark)
Laplace: v = √(γp/ρ) = √(1.4 × 78 295) = √109 613 = 331 m s⁻¹. (1 mark)
Ratio = √γ = √1.4 = 1.18. (1 mark)
Laplace's value agrees with the measured speed because the compressions are adiabatic. (1 mark)

Answer: Newton 280 m s⁻¹; Laplace 331 m s⁻¹; ratio 1.18.

Example 4: Echo and temperature · 3 marks · Short answerHarder board level

The speed of sound at 0 °C is 331 m s⁻¹. Find its value at 27 °C and the distance of a cliff if an echo returns after 3.0 s at that temperature.

Try it first, then show the solutionSolution

v ∝ √T (kelvin): v = 331 × √(300/273). (1 mark)
√(1.0989) = 1.0483, so v = 347 m s⁻¹. (1 mark)
d = vt/2 = 347 × 3/2 = 520 m. (1 mark)

Answer: v = 347 m s⁻¹; d ≈ 520 m.

Example 5: Pulse on a string · 5 marks · Long answerHOTS / competency

A string of mass 5 g and length 2 m is stretched with a tension of 80 N. Find the wave speed, the frequency of a wave of wavelength 0.5 m, the time for a pulse to run the length of the string, and the new speed if the tension is quadrupled.

Try it first, then show the solutionSolution

μ = 0.005/2 = 2.5 × 10⁻³ kg m⁻¹. (1 mark)
v = √(T/μ) = √(80/0.0025) = √32 000 = 179 m s⁻¹. (1 mark)
f = v/λ = 179/0.5 = 358 Hz. (1 mark)
t = L/v = 2/179 = 0.0112 s. (1 mark)
T → 4T doubles the speed: 358 m s⁻¹. (1 mark)

Answer: v = 179 m s⁻¹; f = 358 Hz; t = 11.2 ms; new speed 358 m s⁻¹.

Where students lose marks in Progressive Waves and Speed of Sound

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong sign in the equation
Reading the direction of travel from the sign of kx incorrectly.
Do this: y = A sin(ωt − kx) moves towards +x; y = A sin(ωt + kx) moves towards −x.
Usually costs: 1 mark.
Mistake: Echo distance
Not dividing by 2.
Do this: Sound travels to the obstacle and back: d = vt/2.
Usually costs: 2 marks.
Mistake: Celsius in v ∝ √T
Using 27 and 0 in the ratio.
Do this: Use kelvin: √(300/273).
Usually costs: 2 marks.

Exam tips

Compare the given equation with the standard form term by term.
Convert units: g to kg, cm to m, kHz to Hz.
Explain why Newton's formula fails and Laplace's corrects it.
State that sound is longitudinal and needs a medium.

Practice questions on Progressive Waves and Speed of Sound

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Sound cannot travel through:
(a) air
(b) water
(c) steel
(d) a vacuum
Hint
Sound is a mechanical wave and needs a medium.
Answer
(d) a vacuum
Q2 · 2 marks · Short answerBoard level
A sound of frequency 500 Hz travels at 340 m s⁻¹. Find its wavelength.
Hint
λ = v/f.
Answer
λ = 340/500 = 0.68 m.
Q3 · 3 marks · Short answerHarder board level
Find the phase difference between two points 0.25 m apart on a 400 Hz sound wave travelling at 340 m s⁻¹.
Hint
λ = v/f, then Δφ = 2πΔx/λ.
Answer
λ = 0.85 m. Δφ = 2π × 0.25/0.85 = 1.85 rad = 106°.
Q4 · 3 marks · Short answerHarder board level
Thunder is heard 4.5 s after the lightning flash. Taking the speed of sound as 340 m s⁻¹ and ignoring the travel time of light, find the distance of the lightning.
Hint
d = vt.
Answer
d = 340 × 4.5 = 1530 m.
Q5 · 3 marks · Short answerHarder board level
A wave of speed 60 m s⁻¹ travels on a string of mass per unit length 0.01 kg m⁻¹. Find the tension.
Hint
T = μv².
Answer
T = 0.01 × 3600 = 36 N.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a bat emits ultrasonic pulses at 40 kHz in air (v = 340 m s⁻¹), and a ship's sonar sends pulses in sea water, where sound travels at 1480 m s⁻¹.
(i) Find the wavelength of the bat's pulse. (1 mark)
(ii) An echo returns to the bat after 0.025 s; find the distance of the object. (1 mark)
(iii) A sonar echo returns after 0.40 s; find the depth of the sea bed. (1 mark)
(iv) Find the wavelength of a 40 kHz pulse in sea water. (1 mark)
Hint
The echo travels the distance twice.
Answer
(i) λ = 340/40 000 = 8.5 mm. (ii) d = 340 × 0.025/2 = 4.25 m. (iii) d = 1480 × 0.40/2 = 296 m. (iv) λ = 1480/40 000 = 37 mm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Relation between v, f and λ?Show answer
v = fλ.
2.Speed of a wave on a string?Show answer
√(T/μ).
3.Laplace's formula?Show answer
v = √(γp/ρ).
4.How does the speed of sound change with temperature?Show answer
v ∝ √T (in kelvin).

Progressive Waves and Speed of Sound: frequently asked questions

Does the speed of sound depend on pressure?

For an ideal gas at constant temperature, no: p and ρ change in proportion, so p/ρ stays constant.

Why is the speed of sound greater in a solid?

The elastic modulus of solids is very large compared with the density, and v = √(modulus/density).

What is the difference between ultrasonic and supersonic?

Ultrasonic refers to frequencies above the audible range; supersonic refers to a speed greater than the speed of sound.