1. Physics
  2. Oscillations and Waves
  3. Wave Motion

Oscillations and Waves · JEE & NEET Physics

Wave Motion: notes and previous year questions

Kinds of waves, λ, f and v = fλ, y = A sin(kx − ωt), wave speed in different media, intensity and reflection.

Wave Motion in short

  • A wave is a travelling disturbance that carries energy, not matter.
  • Mechanical waves need a medium; electromagnetic waves do not and travel at c in a vacuum.
  • Transverse: medium moves across the wave; longitudinal: along it, with compressions and rarefactions.
  • In sound, pressure is largest where displacement is zero: a quarter wave apart.

1What is a wave?

Shake the end of a rope and a wave runs along it, but each bit of rope only moves up and down about its own place. A wave is a disturbance that travels, carrying energy from place to place without carrying the medium along.

  • A leaf on a pond bobs as ripples pass; it is not carried across.
  • Sound crossing a room: the air only moves back and forth a tiny distance; the disturbance crosses the room.

2Kinds of waves

Mechanical waves (sound, water waves, seismic waves) need a medium and cannot cross a vacuum. Electromagnetic waves (light, radio, X-rays) need no medium; in a vacuum they all travel at c=3×108c = 3 \times 10^8 m/s. So we see lightning before we hear thunder, and we can see the Sun but not hear it.

In a transverse wave the medium moves at right angles to the wave (a string, light). In a longitudinal wave it moves along the wave, making compressions (crowded) and rarefactions (spread out): sound in air, a slinky pushed in and out, ultrasound in water.

FeatureTransverseLongitudinal
Medium movesat right angles to the wavealong the wave
Patterncrests and troughscompressions and rarefactions
In solidsyesyes
Inside liquids and gasesno (only along the surface)yes
Polarisationyesno

For sound we can graph the displacement s of each air layer, or the extra pressure ΔP. The pressure is largest at a compression, where the layers on both sides have moved in and the displacement itself is zero: the pressure wave is a quarter wave (90°) out of step with the displacement wave.

3Describing a wave

  • Wavelength λ\lambda: the distance between neighbouring points in the same step, such as crest to crest (m).
  • Amplitude AA: the largest displacement from the middle. The energy carried grows as A2A^2.
  • Period TT: the time for one full cycle of any particle, which does SHM.
  • Frequency f=1/Tf = 1/T: waves passing each second (Hz); ω=2πf\omega = 2\pi f.

In one period the wave moves forward exactly one wavelength, so:

v=λT=fλv = \frac{\lambda}{T} = f\lambda

4The wave equation

y(x,t)=Asin⁡(kx−ωt+ϕ)y(x,t) = A\sin(kx - \omega t + \phi)
k=2πλ,ω=2πf,v=ωkk = \frac{2\pi}{\lambda}, \quad \omega = 2\pi f, \quad v = \frac{\omega}{k}k, the wave number (rad/m): the phase changes by 2π over each wavelength.

Which way? At a crest the phase kx−ωtkx - \omega t stays fixed, so as tt grows, xx grows: kx−ωtkx - \omega t moves toward +x+x. With kx+ωtkx + \omega t, xx shrinks as tt grows: the wave moves toward −x-x. Any shape f(x−vt)f(x - vt) moves toward +x+x at speed vv, and g(x+vt)g(x + vt) toward −x-x; both satisfy the wave equation ∂2y/∂t2=v2 ∂2y/∂x2\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2.

Δϕ=2πλ Δx=k Δx\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x = k\,\Delta xPhase difference between two points Δx apart: λ/4 gives π/2, λ/2 gives π.
Path differencePhase difference
0,λ,2λ,…0, \lambda, 2\lambda, \ldots0,2π,4π,…0, 2\pi, 4\pi, \ldots (in step)
λ/2,3λ/2,…\lambda/2, 3\lambda/2, \ldotsπ,3π,…\pi, 3\pi, \ldots (opposite)
λ/4,3λ/4,…\lambda/4, 3\lambda/4, \ldotsπ/2,3π/2,…\pi/2, 3\pi/2, \ldots

Example: k=4k = 4 rad/m and Δϕ=π/3\Delta\phi = \pi/3 give Δx=Δϕ/k=π/12≈0.26\Delta x = \Delta\phi/k = \pi/12 \approx 0.26 m (here λ=π/2\lambda = \pi/2 m).

5Particle velocity and wave velocity

vparticle=∂y∂t=−Aωcos⁡(kx−ωt)v_{particle} = \frac{\partial y}{\partial t} = -A\omega\cos(kx - \omega t)

Each particle moves up and down with a changing velocity: largest, AωA\omega, as it passes the middle, and zero at a crest or trough. The wave velocity ω/k\omega/k is fixed by the medium. For a wave moving right, a particle just ahead of an arriving crest is moving up. Example: y=0.02sin⁡(30x−400t)y = 0.02\sin(30x - 400t) has largest particle speed 0.02×400=80.02 \times 400 = 8 m/s but wave speed 400/30≈13.3400/30 \approx 13.3 m/s.

6Wave speed in different media

The speed is set by the medium: a restoring stiffness over an inertia.

v=Tμv = \sqrt{\frac{T}{\mu}}String: tension T (N), mass per metre μ (kg/m). Tighter is faster; heavier is slower.
v=γRTMv = \sqrt{\frac{\gamma RT}{M}}Gas (γ = 1.4 for air). Grows with the kelvin temperature, not with pressure; faster in light gases.
v=Yρ,v=Bρv = \sqrt{\frac{Y}{\rho}}, \quad v = \sqrt{\frac{B}{\rho}}Solid rod (Young's modulus Y) and liquid (bulk modulus B).

In air, v≈331+0.6 tv \approx 331 + 0.6\,t m/s with tt in °C: about 343 m/s at 20 °C.

MediumSpeed of sound
Air, 20 °Cabout 343 m/s
Waterabout 1480 m/s
Steelabout 5000 m/s
Aluminiumabout 6400 m/s

7Energy, power and intensity

Each metre of string holds on average energy 12μω2A2\tfrac{1}{2}\mu\omega^2A^2, carried forward at speed vv:

P=12μvω2A2P = \tfrac{1}{2}\mu v\omega^2 A^2P ∝ A², P ∝ ω², P ∝ v. For sound use the density ρ and the area.
I=Parea,I=P4πr2I = \frac{P}{\text{area}}, \quad I = \frac{P}{4\pi r^2}Intensity in W/m²; a small source spreads its power over a sphere: the inverse square law.

8Reflection and transmission

  • Fixed end: the pulse comes back upside down, a phase change of π\pi.
  • Free end (a ring sliding on a pole): it comes back the right way up, with no phase change.
  • In both, the amplitude stays the same.

At a join between media with speeds v1v_1 (incoming side) and v2v_2:

Ar=v2−v1v2+v1 Ai,At=2v2v1+v2 AiA_r = \frac{v_2 - v_1}{v_2 + v_1}\,A_i, \quad A_t = \frac{2v_2}{v_1 + v_2}\,A_i

Into a heavier, slower string (v2=v1/2v_2 = v_1/2, e.g. μ2=4μ1\mu_2 = 4\mu_1 at the same tension): Ar=−A/3A_r = -A/3 (inverted) and At=2A/3A_t = 2A/3. Into a lighter, faster string the reflection is upright. The frequency stays the same, so the wavelength changes with the speed: from air into water (about 4.4 times faster) λ\lambda becomes about 4.4 times longer.

Summary

Key ideas

  • A wave is a travelling disturbance that carries energy, not matter.
  • Mechanical waves need a medium; electromagnetic waves do not and travel at c in a vacuum.
  • Transverse: medium moves across the wave; longitudinal: along it, with compressions and rarefactions.
  • In sound, pressure is largest where displacement is zero: a quarter wave apart.
  • In one period a wave moves one wavelength, so v = fλ.
  • y = A sin(kx − ωt + φ) moves toward +x; with kx + ωt it moves toward −x; v = ω/k.
  • Two points Δx apart differ in phase by (2π/λ)Δx.
  • Particle speed (largest Aω) and wave speed (ω/k) are different things.
  • The medium sets the speed: √(T/μ) on strings, √(γRT/M) in gases, √(Y/ρ) in solids, √(B/ρ) in liquids.
  • In a new medium f stays the same and λ changes with v.
  • Power ∝ A²ω²v; from a small source the intensity falls as 1/r².
  • A fixed end inverts a reflected pulse; a free end does not.

Every equation

Wave speed
v=fλ=λ/Tv = f\lambda = \lambda/T
Travelling wave
y=Asin⁡(kx−ωt+ϕ)y = A\sin(kx - \omega t + \phi)
Wave number
k=2π/λk = 2\pi/\lambda
Angular frequency
ω=2πf\omega = 2\pi f
Speed from ω and k
v=ω/kv = \omega/k
Wave equation
∂2y/∂t2=v2 ∂2y/∂x2\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2
Phase difference
Δϕ=(2π/λ) Δx\Delta\phi = (2\pi/\lambda)\,\Delta x
Particle velocity
∂y/∂t=−Aωcos⁡(kx−ωt)\partial y/\partial t = -A\omega\cos(kx - \omega t)
String
v=T/μv = \sqrt{T/\mu}
Gas
v=γRT/Mv = \sqrt{\gamma RT/M}
Air
v≈331+0.6 tv \approx 331 + 0.6\,t
Solid rod
v=Y/ρv = \sqrt{Y/\rho}
Liquid
v=B/ρv = \sqrt{B/\rho}
New medium
λ2/λ1=v2/v1\lambda_2/\lambda_1 = v_2/v_1
Power
P=12μvω2A2P = \tfrac{1}{2}\mu v\omega^2A^2
Point source
I=P/4πr2I = P/4\pi r^2
Reflected amplitude
Ar=v2−v1v2+v1AiA_r = \tfrac{v_2 - v_1}{v_2 + v_1}A_i
Transmitted amplitude
At=2v2v1+v2AiA_t = \tfrac{2v_2}{v_1 + v_2}A_i

Previous year questions with solutions

Real JEE and NEET questions on wave motion. Try each one before you open the solution.

Q1JEE Main 2026Numerical answer

A transverse wave on a string is described by y=3sin⁡(36t+0.018x+π/4)y=3\sin (36t+0.018x+\pi /4). where x,yx,y are in cm and tt in seconds. The least distance between the two successive crests in the wave is ____\_\_\_\_ cm . (Nearest integer)

(π=3.14)(\pi =3.14)

Show answer and solution

Answer: 349

Here xx is in cm, so the number beside xx is k=0.018 rad/cmk = 0.018\ \mathrm{rad/cm}. Two successive crests are one wavelength apart:

λ=2πk=6.280.018≈348.9 cm\lambda = \dfrac{2\pi}{k} = \dfrac{6.28}{0.018} \approx 348.9\ \mathrm{cm}, which is 349 cm349\ \mathrm{cm} to the nearest integer.

The plus sign only says the wave travels towards −x-x, and the phase π4\dfrac{\pi}{4} only shifts where the crests are; neither changes their spacing. The trap is 2π36≈0.17\dfrac{2\pi}{36} \approx 0.17, which uses ω\omega: that is the time between crests passing one point (the period, in seconds), not their distance apart.

Q2NEET 2026One correct option

For a travelling harmonic wave

y(x,t)=2.0cos⁡2π(10t−0.0080x+0.35)y(x,t)=2.0\cos 2\pi (10t-0.0080x+0.35), where xx and yy are in cm and tt in ss. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:

  1. A0.08πrad0.08\pi \mathrm{rad}
  2. B0.8πrad0.8\pi \mathrm{rad}
  3. C8πrad8\pi \mathrm{rad}
  4. D0.008πrad0.008\pi \mathrm{rad}
Show answer and solution

Answer: Option B

Multiply out the 2π2\pi: the phase is 20πt−2π×0.0080x+0.7π20\pi t - 2\pi \times 0.0080x + 0.7\pi, so k=2π×0.0080=0.016π rad/cmk = 2\pi \times 0.0080 = 0.016\pi\ \mathrm{rad/cm}, since xx is in cm. The two points are 0.5 m=50 cm0.5\ \mathrm{m} = 50\ \mathrm{cm} apart, so

Δϕ=k Δx=0.016π×50=0.8π rad\Delta\phi = k\,\Delta x = 0.016\pi \times 50 = 0.8\pi\ \mathrm{rad}

That is B. The time term and the constant 0.350.35 are the same for both points and drop out.

The trap is D, 0.008π0.008\pi, which puts Δx=0.5\Delta x = 0.5 straight in although xx is in centimetres. A, 0.08π0.08\pi, and C, 8π8\pi, are the same slip by other powers of ten (5 cm5\ \mathrm{cm} and 500 cm500\ \mathrm{cm}).

Q3JEE Main 2026One correct option

The equation of a plane progressive wave is given by y=5cos⁡π(200t−x150)y=5\cos \pi (200t-\frac{x}{150}) where xx and yy are in cm and tt is in second. The velocity of the wave is ________ m/s.

  1. A120
  2. B150
  3. C200
  4. D300
Show answer and solution

Answer: Option D

Multiply out the π\pi: y=5cos⁡(200πt−πx150)y = 5\cos\left(200\pi t - \dfrac{\pi x}{150}\right), so ω=200π rad/s\omega = 200\pi\ \mathrm{rad/s} and k=π150 rad/cmk = \dfrac{\pi}{150}\ \mathrm{rad/cm} (since xx is in cm).

v=ωk=200π×150π=30 000 cm/s=300 m/sv = \dfrac{\omega}{k} = 200\pi \times \dfrac{150}{\pi} = 30\,000\ \mathrm{cm/s} = 300\ \mathrm{m/s}

That is D. (Check: f=100 Hzf = 100\ \mathrm{Hz} and λ=2πk=300 cm\lambda = \dfrac{2\pi}{k} = 300\ \mathrm{cm}, and fλ=300 m/sf\lambda = 300\ \mathrm{m/s}.)

The trap is B, 150150, which reads the number under xx as the speed, or C, 200200, which reads the number beside tt as the speed: each is only half the story, and the speed is their product here. A, 120120, fits no reading of the equation.

Practice questions, easy to hard

Three questions from the wave motion practice ladder: one easy, one medium, one hard.

Q4One correct option

Look along a transverse wave on a string at one instant. Crests (the highest points) and troughs (the lowest) repeat at equal spacing. The wavelength λ\lambda is the distance from one crest to the next, or in general between two neighbouring particles in the same state of motion.

In one period TT each particle makes one full oscillation, and in that same time the pattern moves on by exactly one wavelength. So the wave speed is

v=λT=fλv = \dfrac{\lambda}{T} = f\lambda

The speed is fixed by the medium. The frequency is fixed by the source, so when a wave passes from one medium into another its frequency does not change. Its speed does, and the wavelength λ=vf\lambda = \dfrac{v}{f} changes in the same ratio as the speed.

A 500 Hz500\ \mathrm{Hz} tuning fork sends sound through air, where it travels at 340 m/s340\ \mathrm{m/s}, into a lake, where it travels at 1500 m/s1500\ \mathrm{m/s}. What is the wavelength of the sound in the water?

  1. A0.68 m0.68\ \mathrm{m}
  2. B3 m3\ \mathrm{m}
  3. C0.33 m0.33\ \mathrm{m}
  4. D7.5×105 m7.5 \times 10^{5}\ \mathrm{m}
Show answer and solution

Answer: Option B

The frequency stays 500 Hz500\ \mathrm{Hz} in the water, so λ=vf=1500500=3 m\lambda = \dfrac{v}{f} = \dfrac{1500}{500} = 3\ \mathrm{m}.

The trap is A, 0.68 m0.68\ \mathrm{m}, the wavelength in the AIR (340500\dfrac{340}{500}): the wavelength changes with the speed. C, 0.33 m0.33\ \mathrm{m}, is fv\dfrac{f}{v}, the fraction upside down. D multiplies ff by vv, which is not even a length.

Q5One correct option

For a gas, Newton assumed the squeezing happens at a steady temperature, so he used the isothermal bulk modulus, B=pB = p: v=pρv = \sqrt{\dfrac{p}{\rho}}. For air at 0∘C0^{\circ}\mathrm{C} (p=1.01×105 Pap = 1.01 \times 10^{5}\ \mathrm{Pa}, ρ=1.29 kg/m3\rho = 1.29\ \mathrm{kg/m^{3}}) this gives about 280 m/s280\ \mathrm{m/s}, well below the measured 332 m/s332\ \mathrm{m/s}.

Laplace saw the mistake. The compressions and rarefactions come and go so fast that heat has no time to flow between them, so the changes are adiabatic, and the adiabatic bulk modulus is B=γpB = \gamma p. So

v=γpρv = \sqrt{\dfrac{\gamma p}{\rho}}

For air, γ=1.4\gamma = 1.4. By what factor does Laplace's correction multiply Newton's value?

  1. A1.41.4
  2. B0.850.85
  3. C1.961.96
  4. Dabout 1.181.18, that is 1.4\sqrt{1.4}
Show answer and solution

Answer: Option D

The ratio is γp/ρp/ρ=γ=1.4≈1.18\sqrt{\dfrac{\gamma p/\rho}{p/\rho}} = \sqrt{\gamma} = \sqrt{1.4} \approx 1.18, and 280×1.18≈331 m/s280 \times 1.18 \approx 331\ \mathrm{m/s}, which matches the measurement.

The trap is A, 1.41.4: γ\gamma multiplies the modulus, which sits under the square root. C, 1.96=1.421.96 = 1.4^{2}, squares instead of rooting, and B, 11.4\dfrac{1}{\sqrt{1.4}}, goes the wrong way: the adiabatic gas is stiffer, so sound is faster, not slower.

Q6Numerical answer

A girl claps her hands in front of a tall cliff and hears the echo 1.2 s1.2\ \mathrm{s} later. The air is at 27∘C27^{\circ}\mathrm{C}, and sound travels at 332 m/s332\ \mathrm{m/s} in air at 0∘C0^{\circ}\mathrm{C}. Using v∝Tv \propto \sqrt{T}, how far away is the cliff, in metres, to the nearest metre?

Show answer and solution

Answer: 209 m

At 27∘C27^{\circ}\mathrm{C}, T=300 KT = 300\ \mathrm{K}, so v=332300273≈332×1.048≈348 m/sv = 332\sqrt{\dfrac{300}{273}} \approx 332 \times 1.048 \approx 348\ \mathrm{m/s} (the handy rule v0+0.61tv_{0} + 0.61t gives 348.5 m/s348.5\ \mathrm{m/s}, almost the same).

The sound goes there and back, so d=vt2=348×1.22≈209 md = \dfrac{vt}{2} = \dfrac{348 \times 1.2}{2} \approx 209\ \mathrm{m}.

The trap is about 418 m418\ \mathrm{m}, the whole path there and back. Another is 199 m199\ \mathrm{m}, which forgets the warm air and uses 332 m/s332\ \mathrm{m/s}.