The displacement relation
For the mathematical description of a travelling wave we need a function of both position $x$ and time $t$ that, at every instant, gives the shape of the wave, and at every location describes the motion of the constituent of the medium there. For a sinusoidal travelling wave that function must itself be sinusoidal. Taking the wave to be transverse, with $x$ denoting the position of a medium element and $y$ its displacement from equilibrium, NCERT writes the displacement relation in a progressive wave as
$$y(x,t) = a\,\sin(kx - \omega t + \varphi)$$This is the central result of Section 14.3. The Summary of the chapter states the same equation and names the symbols verbatim: $a$ is the amplitude of the wave, $k$ is the angular wave number, $\omega$ is the angular frequency, $(kx-\omega t+\varphi)$ is the phase, and $\varphi$ is the phase constant or phase angle. The presence of $\varphi$ is equivalent to writing the disturbance as a linear combination of a sine and a cosine, $y = A\sin(kx-\omega t) + B\cos(kx-\omega t)$, with $a=\sqrt{A^2+B^2}$ and $\tan\varphi = -B/A$.
Two ways to read the wave
The single equation carries two complementary pictures, and recognising both is the key to interpreting every symbol that follows.
| View | What is held fixed | Reduced argument | What you see |
|---|---|---|---|
| Snapshot in space | time, $t=t_0$ | kx + constant | Shape of the whole wave — a sine curve in $x$ |
| Motion in time | location, $x=x_0$ | −ωt + constant | One medium element executing SHM in $t$ |
As $t$ increases, $x$ must increase to keep $(kx-\omega t+\varphi)$ constant — so the pattern advances along $+x$. A crest, marked as a fixed phase point, moves forward by one wavelength in the time one element completes a full oscillation. That synthesis of the spatial and temporal views is exactly what makes the function a progressive wave rather than a static curve or a stationary oscillation.
Amplitude and phase
Since the sine varies between $+1$ and $-1$, the displacement $y(x,t)$ varies between $+a$ and $-a$. Taking $a$ as a positive constant, it represents the maximum displacement of the medium constituents from equilibrium and is the amplitude of the wave. The displacement $y$ may be positive or negative, but $a$ is always positive.
The quantity $(kx-\omega t+\varphi)$ is the phase of the wave. Given the amplitude, the phase alone determines the displacement at any position and any instant. The phase constant $\varphi$ is the value of the phase at $x=0$ and $t=0$, which is why it is also called the initial phase angle. By a suitable choice of origin and initial time it can be set to zero, so NCERT often drops it and works with $y=a\sin(kx-\omega t)$.
Phase is not the phase constant
Questions on "the phase at the origin" or "initial phase" want the constant $\varphi$ only. Questions on "the phase of the wave" want the entire argument $(kx-\omega t+\varphi)$, which depends on where and when you look. Confusing the two is a frequent slip in equation-reading items.
Rule: phase $=(kx-\omega t+\varphi)$ — varies with $x$ and $t$. Phase constant $=\varphi$ — its value only at $x=0,\ t=0$.
Wavelength and angular wave number
The wavelength $\lambda$ is the minimum distance between two points having the same phase — conveniently, the distance between two consecutive crests or two consecutive troughs. Putting $\varphi=0$, the snapshot at $t=0$ is $y(x,0)=a\sin kx$. Because the sine repeats after every $2\pi$ change in its argument, the displacement repeats when $kx$ increases by $2\pi$, i.e. when $x$ increases by $2\pi/k$. The least such distance gives
$$k\lambda = 2\pi \qquad\Longrightarrow\qquad k = \frac{2\pi}{\lambda}$$Here $k$ is the angular wave number (or propagation constant); its SI unit is radian per metre, $\mathrm{rad\,m^{-1}}$. It represents $2\pi$ times the number of waves accommodated per unit length.
k is 2π/λ, not 1/λ
The angular wave number carries a factor of $2\pi$. The reciprocal $1/\lambda$ is the spatial frequency (waves per metre) and is smaller than $k$ by exactly $2\pi$. Plugging $1/\lambda$ where $k$ is required, or forgetting the $2\pi$ when converting between $k$ and $\lambda$, is the single most common arithmetic error in this subtopic.
Rule: $k=\dfrac{2\pi}{\lambda}\ \mathrm{rad\,m^{-1}}$ and $\lambda=\dfrac{2\pi}{k}$.
Period, angular frequency and frequency
Now fix a location, say $x=0$ with $\varphi=0$, so that $y(0,t)=-a\sin\omega t$. This describes the displacement of one element of the medium as a function of time — a simple harmonic oscillation. The period $T$ is the time an element takes to complete one full oscillation. Since the sine repeats after every $2\pi$, demanding that the motion repeat after $T$ gives $\omega T = 2\pi$, hence
$$\omega = \frac{2\pi}{T}, \qquad \nu = \frac{1}{T} = \frac{\omega}{2\pi}$$Here $\omega$ is the angular frequency with SI unit $\mathrm{rad\,s^{-1}}$, and $\nu$ is the frequency — the number of oscillations per second, measured in hertz. For a longitudinal wave the same relation holds with the displacement $y$ replaced by $s(x,t)=a\sin(kx-\omega t+\varphi)$, where $s$ is measured along the direction of propagation.
Figures 1 and 2 look almost identical, yet they are physically distinct. The first is a photograph of space at one instant, with the repeat distance $\lambda$; the second is a record of one point through time, with the repeat interval $T$. Distinguishing the horizontal axis — distance versus time — is what separates $\lambda$ from $T$ and $k$ from $\omega$.
The parameter set at a glance
The seven quantities of the progressive-wave relation form a tightly linked set. The table collects each symbol with its meaning, defining equation and SI unit.
| Symbol | Name | Defining relation | SI unit |
|---|---|---|---|
a | Amplitude | max value of $|y|$ | m |
k | Angular wave number | $k = 2\pi/\lambda$ | rad m⁻¹ |
λ | Wavelength | $\lambda = 2\pi/k$ | m |
ω | Angular frequency | $\omega = 2\pi/T = 2\pi\nu$ | rad s⁻¹ |
T | Period | $T = 2\pi/\omega$ | s |
ν | Frequency | $\nu = 1/T = \omega/2\pi$ | Hz |
v | Wave speed | $v = \omega/k = \nu\lambda$ | m s⁻¹ |
The last row, $v=\dfrac{\omega}{k}=\nu\lambda$, follows from requiring a fixed phase point ($kx-\omega t=\text{constant}$) to move at constant speed. It is a general relation for all progressive waves: in the time one element completes a full oscillation, the pattern advances by exactly one wavelength. The detailed derivation belongs to the next subtopic.
The result $v=\omega/k=\nu\lambda$ is unpacked, with the fixed-phase-point argument, in Speed of a Travelling Wave.
Direction of travel
The relative sign of the $kx$ and $\omega t$ terms encodes the direction of propagation. The form $y=a\sin(kx-\omega t+\varphi)$, with opposite signs, represents a wave moving along the positive $x$-axis. Replacing $-\omega t$ by $+\omega t$,
$$y(x,t) = a\,\sin(kx + \omega t + \varphi)$$describes a wave travelling in the negative $x$-direction. The reasoning is the constant-phase argument: with $(kx-\omega t)$, holding the phase fixed forces $x$ to grow as $t$ grows, so the pattern advances toward $+x$; with $(kx+\omega t)$, $x$ must decrease as $t$ grows, so the pattern moves toward $-x$.
Opposite signs → +x; like signs → −x
Do not be misled by which term is written first. A wave $y=a\sin(\omega t - kx)$ has opposite signs on $kx$ and $\omega t$ and so still travels along $+x$; $y=a\sin(\omega t + kx)$ travels along $-x$. The decisive feature is the relative sign, not the order of terms — and an overall minus sign in front of the amplitude only flips the wave upside down, it does not change the direction.
Rule: $kx$ and $\omega t$ with opposite signs ⇒ travels along $+x$; with the same sign ⇒ travels along $-x$.
Reading values off an equation
The standard NEET task is to compare a given equation, term by term, with $y=a\sin(kx-\omega t+\varphi)$ and extract the parameters. The two worked examples below follow the NCERT method exactly.
A wave on a string is $y(x,t)=0.005\,\sin(80.0\,x - 3.0\,t)$, with constants in SI units. Find the amplitude, wavelength, period and frequency.
Compare with $y=a\sin(kx-\omega t)$. Then $a=0.005\ \text{m}=5\ \text{mm}$, $k=80.0\ \text{rad m}^{-1}$ and $\omega=3.0\ \text{rad s}^{-1}$.
$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{80.0} \approx 7.85\ \text{cm}$.
$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{3.0} \approx 2.09\ \text{s}$, and $\nu = \dfrac{1}{T} \approx 0.48\ \text{Hz}$.
A transverse wave is $y(x,t)=3.0\,\sin(36\,t + 0.018\,x + \pi/4)$, with $x,y$ in cm and $t$ in s. State its direction, speed, amplitude, frequency and initial phase.
The terms $36\,t$ and $0.018\,x$ have the same sign, so the wave travels along the negative $x$-direction. Matching $\omega t + kx$ gives $\omega = 36\ \text{rad s}^{-1}$ and $k = 0.018\ \text{rad cm}^{-1}$.
Speed $v = \dfrac{\omega}{k} = \dfrac{36}{0.018} = 2000\ \text{cm s}^{-1} = 20\ \text{m s}^{-1}$.
Amplitude $a = 3.0\ \text{cm}$; frequency $\nu = \dfrac{\omega}{2\pi} = \dfrac{36}{2\pi} \approx 5.7\ \text{Hz}$; initial phase at the origin $\varphi = \pi/4$.
Displacement Relation — the essentials
- A harmonic progressive wave along $+x$ is $y(x,t)=a\sin(kx-\omega t+\varphi)$; the $+\omega t$ form travels along $-x$.
- Amplitude $a$ is the maximum $|y|$ and is always positive; phase is the full argument $(kx-\omega t+\varphi)$; phase constant $\varphi$ is its value at $x=0,\ t=0$.
- Angular wave number $k=2\pi/\lambda$ (rad m⁻¹) — never $1/\lambda$.
- Angular frequency $\omega=2\pi/T=2\pi\nu$ (rad s⁻¹); period $T=2\pi/\omega$; frequency $\nu=1/T$.
- Wave speed $v=\omega/k=\nu\lambda$ for all progressive waves.
- To analyse a given equation: match it to the standard form, read $a$, $k$, $\omega$, then derive $\lambda$, $T$, $\nu$, $v$ and the direction from the relative sign.