MHT-CET Physics · Superposition of Waves

Progressive Waves: the Wave Equation, Phase and Particle Velocity

A progressive wave y = A sin(ωt − kx) carries a disturbance at v = ω/k = fλ; two points Δx apart differ in phase by 2πΔx/λ, and each particle oscillates with a top speed Aω that is quite separate from the wave's speed.

All Superposition of Waves notes MHT-CET Physics strategy

Why this matters

34 PYQs, three HARD. Three shapes: reading ω, k, λ, v and the direction off a wave equation (and the speed on a string, √(T/μ)), converting between phase and path difference, and comparing the particles' maximum velocity with the wave velocity.

Concept 1 of 3: Reading the Wave Equation

Everything is in the two coefficients: the number multiplying t is ω, the one multiplying x is k. Their ratio is the speed, and the relative sign of the t and x terms gives the direction — opposite signs, the wave moves toward +x; same signs, toward −x.

Definition

$y = A\sin(\omega t - kx + \phi)$ : $f = \dfrac{\omega}{2\pi}$ , $\lambda = \dfrac{2\pi}{k}$ , $v = \dfrac{\omega}{k} = f\lambda$ . Opposite signs: along $+x$ ; same signs: along $-x$ .
$y = A\sin 2\pi\left(\dfrac{t}{T} - \dfrac{x}{\lambda}\right)$ reads off $T$ and $\lambda$ directly.
Frequency is fixed by the source; entering a new medium changes $v$ and $\lambda$ together.
String: $v = \sqrt{\dfrac{T}{\mu}}$ . Hanging rope with a load: tension, and so $\lambda$ , grows toward the top.
Sound: $v \propto \sqrt{T_{\text{abs}}}$ . Reflection at a rigid wall: phase reverses ( $180^\circ$ ), speed unchanged.

Progressive wave

y = A\sin(\omega t - kx), \qquad v = \frac{\omega}{k} = f\lambda

Worked example

y = 0.05\sin(200\pi t - 4\pi x)

(SI). Frequency, wavelength, speed and direction?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q49Easy

Example 1 · Superposition of Waves · Progressive Wave, Wave Equation, and Velocity

The equation of wave is

Y=6\sin\!\left(12\pi t-0.02\pi x+\frac{\pi}{3}\right)

where '

x

' is in m and '

t

' in s. The velocity of the wave is

Leaving π in the wave number

12\pi t - 0.02\pi x

k = 0.02\pi

, so

v = 600

m/s. Reading

k = 0.02

gives

600\pi

— watch whether the

\pi

is attached to x.

Drill 13 more on reading the wave equation

Concept 2 of 3: Phase Difference and Path Difference

One wavelength of distance is one full cycle of phase, 2π. So any separation along the wave converts to phase by the fraction of a wavelength it covers — 60° is a sixth of a wavelength.

Definition

$\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x$ , $\Delta x = \dfrac{\lambda}{2\pi}\Delta\phi$ .
Points $60^\circ$ apart: $\dfrac{\lambda}{6}$ ; $90^\circ$ : $\dfrac{\lambda}{4}$ .
Two waves at a point $(x, t)$ : phase difference = difference of their full phases, e.g. $(25x - 40t) - (20x - 30t)$ .
$a\cos(\theta)$ leads $a\sin(\theta)$ by $\dfrac{\pi}{2}$ , so a sine and a cosine wave with an extra $\phi$ differ by $\phi + \dfrac{\pi}{2}$ .
Points in the same state of vibration are a whole number of wavelengths apart.

Phase and path

\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x

Worked example

Sound of 680 Hz in air at 340 m/s. How far apart are two points

60^\circ

out of phase?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q19Easy

Example 2 · Superposition of Waves · Progressive Wave, Wave Equation, and Velocity

A transverse wave travelling along a stretched string has a speed of 30 m/s and a frequency of 250 Hz. The phase difference between two points on the string 10 cm apart at the same instant is

Mixing degrees and radians

\Delta x = \frac{\lambda}{2\pi}\Delta\phi

needs

\Delta\phi

in radians; in degrees use

\frac{\Delta\phi}{360^\circ}\lambda

. Plugging 60 into the radian form is the classic slip.

Drill 12 more on phase difference and path difference

Concept 3 of 3: Particle Velocity Against Wave Velocity

The wave moves along x at ω/k; each particle just bobs up and down, fastest as it passes the middle, at Aω. Their ratio, Aω ÷ (ω/k) = Ak, depends only on the amplitude and the wavelength.

Definition

Particle: $v_{p,\max} = A\omega$ , $a_{\max} = A\omega^2$ . Wave: $v = \dfrac{\omega}{k}$ .
$\dfrac{v_{p,\max}}{v} = Ak = \dfrac{2\pi A}{\lambda}$ .
$y = a\sin 2\pi(bt - cx)$ : $v = \dfrac{b}{c}$ , $v_{p,\max} = 2\pi ab$ .
$y = A\sin^2(\omega t - kx)$ is a wave of amplitude $\dfrac{A}{2}$ at $2\omega$ and $2k$ : $v_{p,\max} = A\omega$ , wavelength $\dfrac{\pi}{k}$ .
Energy of a wave $\propto n^2A^2$ .

Speed ratio

\frac{v_{p,\max}}{v_{\text{wave}}} = Ak = \frac{2\pi A}{\lambda}

Worked example

y = 0.02\sin(100t - 5x)

(SI). Maximum particle speed, wave speed and their ratio?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q45Moderate

Example 3 · Superposition of Waves · Progressive Wave, Wave Equation, and Velocity

The equation of a progressive wave is

Y=a\sin 2\pi\!\left(nt-\frac{x}{5}\right)

. The ratio of maximum particle velocity to wave velocity is

Treating a sin² wave like a sine wave

A\sin^2\theta = \frac{A}{2}(1 - \cos 2\theta)

: the oscillation has amplitude

\frac{A}{2}

at twice the frequency, so the particles peak at

A\omega

, not

2A\omega

— and the wavelength is

\frac{\pi}{k}

Drill 6 more on particle velocity against wave velocity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Reading the Wave Equation
Progressive wave
$y = A\sin(\omega t - kx), \qquad v = \frac{\omega}{k} = f\lambda$
Phase Difference and Path Difference
Phase and path
$\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x$
Particle Velocity Against Wave Velocity
Speed ratio
$\frac{v_{p,\max}}{v_{\text{wave}}} = Ak = \frac{2\pi A}{\lambda}$

Watch out for (3)

Leaving π in the wave number
→ Reading the Wave Equation
Mixing degrees and radians
→ Phase Difference and Path Difference
Treating a sin² wave like a sine wave
→ Particle Velocity Against Wave Velocity

Test yourself on Superposition of Waves

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Superposition of Waves chapter test Drill all 34 questions

Related notes

Stationary Waves and Strings →