PhysicsCBSE Class 11 · Waves

Progressive Wave Equation and Speed of Waves Class 11

Updated 17 October 2026 · Fact-checked

A wave travelling along +x is y = A sin(kx − ωt + φ), with k = 2π/λ, ω = 2πf and speed v = ω/k = fλ. The speed on a stretched string is v = √(T/μ) and the speed of sound in a gas is v = √(γP/ρ) = √(γRT/M), which rises as √T. Frequency is fixed by the source; wavelength changes with the medium.

Understand Progressive Wave Equation and Speed of Waves

Types of wave. In a transverse wave the particles of the medium move at right angles to the direction of travel (waves on a string); in a longitudinal wave they move along it (sound in air, forming compressions and rarefactions). Waves transfer energy, not matter.

Wave equation. A sinusoidal wave moving in the $+ x$ direction is
$ $y (x, t) = A sin (k x - ω t + ϕ), k = \frac{2 π}{λ}, ω = \frac{2 π}{T} = 2 π f, v = \frac{ω}{k} = f λ .$ $
$k x + ω t$ gives a wave moving towards $- x$ . The phase $k x - ω t$ is constant for a point moving with speed $v$ . Particle velocity is $\partial y / \partial t$ , with maximum $ω A$ , which is different from the wave speed $v$ .

Phase and path difference. Two points separated by $Δ x$ have phase difference $Δ ϕ = k Δ x = \frac{2 π}{λ} Δ x$ ; the same point at times $Δ t$ apart differs by $ω Δ t$ . A path difference of $λ$ is a phase difference of $2 π$ .

Speed in a medium.

*Stretched string:* $v = T / μ$ , where $T$ is the tension and $μ$ the mass per unit length. Doubling the tension raises the speed by $2$ .
*Sound in a gas:* Newton assumed isothermal compressions and gave $v = P / ρ$ , about 15–16% too low (about 280 m/s against about 331 m/s at 0 °C); Laplace corrected this using adiabatic compressions: $v = γ P / ρ = γ R T / M$ . So $v \propto T$ , is independent of pressure (at fixed $T$ , $P / ρ$ is constant), increases with humidity (damp air is less dense), and is greater in lighter gases.
In a given medium the speed depends on the medium only, not on the frequency or amplitude.

At a boundary. Frequency stays the same when a wave moves from one medium to another; speed and wavelength change in proportion.

Progressive Wave Equation and Speed of Waves: formulas and key facts

Travelling wave: $y = A sin (k x - ω t + ϕ)$; Towards +x.
Wave speed: $v = \frac{ω}{k} = f λ$; $k = 2 π / λ$ .
Phase difference: $Δ ϕ = \frac{2 π}{λ} Δ x = ω Δ t$; Space and time.
Speed on a string: $v = \frac{T}{μ}$; $μ$ in kg/m.
Speed of sound in a gas: $v = \frac{γ P}{ρ} = \frac{γ R T}{M}$; Laplace.
Maximum particle speed: $v_{p, ma x} = ω A$; Not the wave speed.

How to solve Progressive Wave Equation and Speed of Waves questions

Compare the given equation with y = A sin(kx − ωt + φ) to read A, k, ω, and then compute f, λ and v.

1Match the equation to y = A sin(kx ∓ ωt + φ) and read A, k and ω.
2Find λ = 2π/k, f = ω/2π and v = ω/k; the sign tells the direction.
3For a string, find μ = m/L and v = √(T/μ).
4For sound, use v = √(γRT/M) or scale v ∝ √T (in kelvin).
5For phase difference, use 2πΔx/λ (points) or ωΔt (instants).
6Remember the frequency does not change at a boundary.

Progressive Wave Equation and Speed of Waves 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.

Pulse in a clothesline

Suppose a clothesline of mass per unit length 0.050 kg/m is under a tension of 200 N.

Find the speed of a pulse and its travel time over 12 m.

Try it first, then show the solutionSolution

v = √(T/μ) = √(200/0.050) = √4000 = 63.2 m/s.
t = 12/63.2 = 0.19 s.

Answer: 63 m/s; 0.19 s.

Memory hook: Tighter or lighter lines carry pulses faster.

Thunder and lightning

Suppose thunder is heard 6.0 s after the flash. Take the speed of sound as 340 m/s and neglect the travel time of light.

Estimate the distance of the lightning.

Try it first, then show the solutionSolution

d = vt = 340 × 6.0 = 2040 m.

Answer: About 2.0 km.

Memory hook: Light arrives almost instantly.

Tricks to remember Progressive Wave Equation and Speed of Waves

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

Read k and ω
Compare with y = A sin(kx − ωt): v = ω/k.
Use it when: Equation questions.
Tension quadruples, speed doubles
v ∝ √T and v ∝ 1/√μ.
Use it when: String problems.
Sound ∝ √T
Use kelvin temperatures.
Use it when: Speed of sound with temperature.
Frequency is the source's
It stays fixed across a boundary.
Use it when: Medium change.

Progressive Wave Equation and Speed of Waves: solved examples

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

Example 1: Reading a wave equation · 2 marks · Very short answerBoard level

A wave on a string is y = 0.08 sin(5x − 200t + π/4) (SI units). Find the amplitude, wavelength, frequency, speed and direction of travel.

Try it first, then show the solutionSolution

A = 0.08 m; k = 5 rad/m so λ = 2π/5 = 1.26 m; ω = 200 rad/s so f = 200/2π = 31.8 Hz. (1)
v = ω/k = 40 m/s; kx − ωt means travel towards +x. (1)

Answer: 0.08 m; 1.26 m; 31.8 Hz; 40 m/s towards +x.

Example 2: Speed on a string · 3 marks · Short answerBoard level

A string of mass 12.5 g and length 2.5 m is stretched under a tension of 80 N. Find the wave speed, the time for a pulse to travel the full length, and the speed if the tension is quadrupled.

Try it first, then show the solutionSolution

μ = 0.0125/2.5 = 5.0 × 10⁻³ kg/m; v = √(80/0.005) = 126.5 m/s. (1)
t = 2.5/126.5 ≈ 0.020 s. (1)
Quadrupling the tension doubles the speed: 253 m/s. (1)

Answer: 126.5 m/s; 0.020 s; 253 m/s.

Example 3: Speed of sound and temperature · 3 marks · Short answerHarder board level

The speed of sound in air at 0 °C is 331 m/s. Find its speed at 27 °C and the temperature at which it is 350 m/s.

Try it first, then show the solutionSolution

v ∝ √T (kelvin): v₂ = 331 × √(300/273) ≈ 347 m/s. (1)
T = 273 × (350/331)² = 273 × 1.118 ≈ 305 K. (1)
That is about 32 °C. (1)

Answer: 347 m/s; about 32 °C.

Example 4: Phase relations · 4 marks · Long answerHarder board level

A sound wave of frequency 170 Hz travels in air at 340 m/s. (a) Find λ, k and ω. (b) Find the phase difference between two points 0.5 m apart along the wave. (c) Find the time interval corresponding to a phase change of π at a fixed point. (d) Write the wave equation for amplitude 1.0 mm.

Try it first, then show the solutionSolution

(a) λ = v/f = 2.0 m; k = 2π/λ = π rad/m; ω = 2π × 170 = 340π rad/s. (1)
(b) Δφ = kΔx = π × 0.5 = π/2. (1)
(c) Δt = Δφ/ω = π/340π = 2.94 ms (half a period). (1)
(d) y = 1.0 × 10⁻³ sin(πx − 340πt) m. (1)

Answer: (a) 2.0 m, π rad/m, 340π rad/s; (b) π/2; (c) 2.94 ms; (d) y = 10⁻³ sin(πx − 340πt).

Example 5: Pulse on a hanging rope · 4 marks · Long answerHOTS / competency

A uniform rope of length 4.0 m and mass M hangs vertically from a support (g = 10 m/s²). Show that the speed of a transverse pulse at height y above the free lower end is √(gy), and find the speed at y = 1 m, at the top, and the time for a pulse to travel from bottom to top.

Try it first, then show the solutionSolution

The tension at height y supports the rope below: T = μgy, so v = √(T/μ) = √(gy). (1)
At y = 1 m: v = √10 = 3.16 m/s; at the top: v = √(40) = 6.32 m/s. (1)
Time: dt = dy/√(gy), so t = ∫dy/√(gy) from 0 to L = 2√(L/g). (1)
t = 2√(4.0/10) = 1.26 s. (1)

Answer: √(gy); 3.16 m/s; 6.32 m/s; 1.26 s.

Where students lose marks in Progressive Wave Equation and Speed of Waves

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

Mistake: Using the particle speed as the wave speed.
Confusing ωA with ω/k.
Do this: v = ω/k is the wave speed.
Usually costs: The answer mark.
Mistake: Using the temperature in °C for v ∝ √T.
Not using kelvin.
Do this: Convert: T(K) = t + 273.
Usually costs: Whole numerical.
Mistake: Saying the frequency changes at a boundary.
Mixing with wavelength.
Do this: Frequency is the source's and fixed.
Usually costs: Concept mark.
Mistake: Using mass instead of mass per unit length.
Wrong μ.
Do this: μ = m/L.
Usually costs: The answer mark.

Exam tips

Write the standard form y = A sin(kx − ωt) and read k and ω.
Give the direction of travel from the relative sign of kx and ωt.
Quote the formula for the speed of sound including γ.
State that the speed in a medium does not depend on frequency.
For phase difference give the answer as a multiple of π.

Practice questions on Progressive Wave Equation and Speed of Waves

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 · MCQHarder board level
In the wave y = A sin(kx − ωt), the maximum speed of a particle of the medium equals the wave speed when
(a) kA = 1 (b) A = λ (c) kA = 2π (d) ωA = 1
Hint
v_p,max = ωA and v = ω/k.
Answer
(a) kA = 1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The speed of a transverse wave on a given stretched string is independent of its frequency.
Reason (R): The speed depends only on the tension and the mass per unit length of the string.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Medium properties set v.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the wavelength of sound of frequencies 20 Hz and 20 kHz in air where the speed is 340 m/s.
Hint
λ = v/f.
Answer
λ = 340/20 = 17 m and 340/20000 = 0.017 m (1.7 cm).
Q4 · 3 marks · Short answerHarder board level
The tension in a string of mass per unit length 0.010 kg/m is raised from 100 N to 144 N. Find the wave speeds and the percentage change in speed.
Hint
v = √(T/μ).
Answer
v₁ = √(100/0.01) = 100 m/s; v₂ = √(144/0.01) = 120 m/s; increase 20%.
Q5 · 4 marks · Long answerHOTS / competency
Two strings of mass per unit length 0.010 kg/m and 0.040 kg/m are joined end to end and held under a tension of 100 N. A 50 Hz source sends a wave from the first into the second. Find the speed, wavelength and wave number in each, and the time for a pulse to cross 5 m of each.
Hint
The frequency is the same in both.
Answer
v₁ = √(100/0.01) = 100 m/s, λ₁ = 2.0 m, k₁ = π rad/m. v₂ = √(100/0.04) = 50 m/s, λ₂ = 1.0 m, k₂ = 2π rad/m. Times: 5/100 = 0.05 s and 5/50 = 0.10 s.

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.Wave speed in terms of f and λ?Show answer
v = fλ.
2.Speed on a string?Show answer
√(T/μ).
3.Speed of sound (Laplace)?Show answer
√(γP/ρ).
4.Phase difference for path difference Δx?Show answer
2πΔx/λ.
5.Which quantity is the same in two media?Show answer
Frequency.
6.Newton's error?Show answer
He used the isothermal bulk modulus instead of the adiabatic one.

Progressive Wave Equation and Speed of Waves: frequently asked questions

Why does sound travel faster on a hot day?

The speed varies as √T, so a higher temperature gives a higher speed.

Does the speed of sound depend on the pressure of the air?

Not at constant temperature: P and ρ change in the same proportion, so P/ρ and the speed stay the same.

What is the difference between wave speed and particle speed?

The wave speed is how fast the disturbance moves through the medium; the particle speed is the oscillation speed of a medium particle, up to ωA.

Why are sound waves longitudinal in air?

Air cannot support shear, so only compressions and rarefactions can travel.

What does the sign in kx − ωt tell you?

kx − ωt describes a wave moving towards +x; kx + ωt moves towards −x.