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NEET UG · Physics

Waves

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NEET UG · Physics

Continue studying Waves

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Find wave direction, wavelength and frequency, distinguish wave speed from particle speed and check phase before NEET Waves MCQs. Travelling waves: separate wave speed from particle motion

About 4 min · NEET UG

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Travelling waves: separate wave speed from particle motion

Find wave direction, wavelength and frequency, distinguish wave speed from particle speed and check phase before NEET Waves MCQs.

What you will understand

Read the units and coefficients in a sinusoidal travelling-wave equation. Distinguish movement of the wave pattern from motion of a material point, then compare points at the same instant.

Quick answer: specify the convention

For y=Asin⁡(kx−ωt+ϕ)y=A\sin(kx-\omega t+\phi) with positive kk and ω\omega, constant phase requires xx to increase as tt increases: the pattern travels toward +x+x with speed ω/k\omega/k. With kx+ωtkx+\omega t, it travels toward −x-x. Here k=2π/λk=2\pi/\lambda and ω=2πf\omega=2\pi f. A material point on the ideal transverse string oscillates in yy; it does not follow the crest along the string.

Follow the pattern and one material point

Two snapshots of y=A sin(kx−omega t), with k and omega positive, show a right-moving wave. At t=0 a crest is at lambda/4; at t=T/4 that crest is at lambda/2. A marked material point stays at x=0 while its transverse displacement changes from zero to minus A. Green marks a crest, orange marks the same material point; the markers represent different things.
Two snapshots of y=A sin(kx−omega t), with k and omega positive, show a right-moving wave. At t=0 a crest is at lambda/4; at t=T/4 that crest is at lambda/2. A marked material point stays at x=0 while its transverse displacement changes from zero to minus A. Green marks a crest, orange marks the same material point; the markers represent different things.

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Worked example: read the coefficients and units

Let y=0.020sin⁡(4πx−40πt)y=0.020\sin(4\pi x-40\pi t), where x,yx,y are in metres and tt is in seconds. The amplitude is 0.020 m0.020\,\mathrm m; k=4π rad m−1k=4\pi\,\mathrm{rad\,m^{-1}} and ω=40π rad s−1\omega=40\pi\,\mathrm{rad\,s^{-1}}. Hence λ=0.50 m\lambda=0.50\,\mathrm m, f=20 Hzf=20\,\mathrm{Hz} and T=0.050 sT=0.050\,\mathrm s. The pattern travels at 10 m s−110\,\mathrm{m\,s^{-1}} toward +x+x. The maximum transverse particle speed is Aω=0.80π m s−1A\omega=0.80\pi\,\mathrm{m\,s^{-1}}, about 2.51 m s−12.51\,\mathrm{m\,s^{-1}}. These are different quantities and directions.

Phase depends on the event being compared

For this one wave, the signed change from event 1 to event 2 is Δθ=kΔx−ωΔt\Delta\theta=k\Delta x-\omega\Delta t. At the same time set Δt=0\Delta t=0. At the same position set Δx=0\Delta x=0. “In phase” means the phase difference is an integer multiple of 2π2\pi. Equal displacement at one instant is insufficient: the two points may be on different parts of the sine cycle.

Compare space and time separately

For one wave y=A sin(kx−omega t), at the same time increasing x by lambda/4 increases phase by pi/2. At the same position increasing time by T/4 decreases phase by pi/2. The top plot is displacement against x/lambda at t=0; the lower plot is displacement against t/T at x=0. A quarter wavelength and a quarter period enter with opposite signs for this convention.
For one wave y=A sin(kx−omega t), at the same time increasing x by lambda/4 increases phase by pi/2. At the same position increasing time by T/4 decreases phase by pi/2. The top plot is displacement against x/lambda at t=0; the lower plot is displacement against t/T at x=0. A quarter wavelength and a quarter period enter with opposite signs for this convention.

Open full-size diagram ↗

Worked example: two changes can cancel

For the numerical wave above, a separation Δx=+0.125 m\Delta x=+0.125\,\mathrm m at the same time gives Δθ=+π/2\Delta\theta=+\pi/2. At one fixed position, a delay Δt=+0.0125 s\Delta t=+0.0125\,\mathrm s gives Δθ=−π/2\Delta\theta=-\pi/2. Now compare two events with both changes together: Δθ=π/2−π/2=0\Delta\theta=\pi/2-\pi/2=0. They lie on the same phase of the right-moving pattern. This does not mean that one material particle moved 0.125 m0.125\,\mathrm m along the string.

Common mistakes to catch

The coefficient kk is not 1/λ1/\lambda, and ω\omega is not frequency in hertz. Include the factors of 2π2\pi. The speed ω/k\omega/k describes propagation; AωA\omega is the maximum speed of a material point in this sinusoidal model. In a transverse string wave the latter direction is perpendicular to propagation. A snapshot against position is different from a displacement–time graph at one position.

Try it yourself 1: identify the speed

For the worked wave, which is the pattern speed? A. 10 m s−110\,\mathrm{m\,s^{-1}}; B. 0.80π m s−10.80\pi\,\mathrm{m\,s^{-1}}; C. 20 m s−120\,\mathrm{m\,s^{-1}}.

Reveal the answer

A. Use ω/k=10 m s−1\omega/k=10\,\mathrm{m\,s^{-1}}. B is the maximum transverse particle speed, and 20 is the numerical value of frequency in hertz, not a speed.

Try it yourself 2: compare at the same instant

Two points of the worked wave are separated by +λ/4+\lambda/4 in the xx direction at the same time. What is the signed phase change? A. 00; B. +π/2+\pi/2; C. −π/2-\pi/2.

Reveal the answer

B. Set Δt=0\Delta t=0 and use kΔx=(2π/λ)(λ/4)=+π/2k\Delta x=(2\pi/\lambda)(\lambda/4)=+\pi/2. The negative quarter-cycle change belongs to a later time at a fixed position in this convention.

A useful check on your next visit

Draw a snapshot and a displacement–time trace for one material point. Label the horizontal axis on each. Recover λ\lambda, ff, pattern speed and maximum particle speed from the worked equation, naming the units before checking your answer.

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Sources and review

Technical review: Codex (automated). Sources checked on 23 October 2026. The diagrams, worked numerical models and self-checks are original teaching material. Human subject review is not claimed.

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