Wave Displacement Functions

From Oscillation to Propagation

You're already familiar with Simple Harmonic Motion (SHM), where an object oscillates back and forth in time around an equilibrium point. A progressive wave takes this one step further. It's an oscillation that also travels through space.

Imagine a long rope. If you flick one end up and down in SHM, you create a pulse that travels down the rope. Each particle of the rope simply moves up and down (oscillating in time), but the wave pattern itself moves forward (propagating in space). To describe this, we need a function that depends on both position ( $x$ ) and time ( $t$ ). This is the wave function, written as $y(x, t)$ .

The function $y(x, t)$ gives the displacement ( $y$ ) of a particle at any position ( $x$ ) and at any instant ( $t$ ).

The Wave Equation

The mathematical description of a simple, one-dimensional sinusoidal wave is a powerful tool. It captures the entire behavior of the wave in a single line.

y(x, t) = A \sin(kx - \omega t + \phi)

The phase contains two crucial new terms: $k$ , the angular wave number, and $\omega$ , the angular frequency.

Angular frequency ( $\omega$ ) should be familiar from SHM. It describes how quickly the wave oscillates in time at a fixed position. A higher $\omega$ means more oscillations per second.
Angular wave number ( $k$ ) is its spatial counterpart. It describes how quickly the wave oscillates in space at a fixed moment in time. A higher $k$ means more oscillations per meter, or a shorter wavelength.

Decoding Wave Parameters

The angular wave number and angular frequency are directly related to the more intuitive properties of wavelength and frequency. Let's see how.

The wavelength, $\lambda$ , is the spatial period of the wave. It's the distance over which the wave's shape repeats. For the sine function to complete one full cycle, its argument (the phase) must change by $2\pi$ . If we hold time constant and move a distance of one wavelength, the phase must change by $2\pi$ solely due to the change in $x$ .

k(x+\lambda) - kx = 2\pi \implies k\lambda = 2\pi

k = \frac{2\pi}{\lambda}

Similarly, the period, $T$ , is the time it takes for one full oscillation at a fixed point. If we stay at a single position ( $x$ is constant) and let time advance by one period, the phase must also change by $2\pi$ .

\omega(t+T) - \omega t = 2\pi \implies \omega T = 2\pi

Since frequency $f$ is the inverse of the period ( $f = 1/T$ ), we get the familiar relationship from SHM:

\omega = 2\pi f

Wave Speed vs. Particle Speed

It's critical to distinguish between two different speeds. The wave speed ( $v$ ) is the speed at which the wave pattern itself propagates. A surfer rides the wave at the wave speed. The particle speed ( $v_p$ ) is the speed of an individual particle in the medium as it oscillates up and down. A duck bobbing on the water moves at the particle speed.

The wave speed is determined by keeping the phase constant. For a point on the wave to maintain its displacement (e.g., stay at a crest), the phase $(kx - \omega t)$ must not change as $x$ and $t$ vary. Therefore, the derivative of the phase with respect to time must be zero.

\frac{d}{dt}(kx - \omega t) = 0 \implies k\frac{dx}{dt} - \omega = 0

Recognizing that $dx/dt$ is the wave speed $v$ , we can solve for it.

v = \frac{dx}{dt} = \frac{\omega}{k}

We can express this in terms of frequency and wavelength:

v = \frac{2\pi f}{2\pi / \lambda} = f\lambda

The particle speed, however, is the velocity of a particle moving in SHM. We find it by taking the partial derivative of the displacement function $y(x,t)$ with respect to time.

v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t + \phi)

Wave speed $v$ is a constant property of the wave, while particle speed $v_p$ is a variable quantity that changes throughout the oscillation.

Now, let's put it all together with a common problem type.

Example Problem

Consider a progressive wave described by the equation: $y(x, t) = 0.05 \sin(2\pi x - 4\pi t)$ where all units are in the SI system.

Our task is to extract all the wave parameters. We compare this to the standard form $y(x, t) = A \sin(kx - \omega t)$ .

Parameter	Comparison	Value
Amplitude ( $A$ )	$A$	$0.05$ m
Angular Wave Number ( $k$ )	$k = 2\pi$	$2\pi$ rad/m
Angular Frequency ( $\omega$ )	$\omega = 4\pi$	$4\pi$ rad/s

From these, we can calculate the wavelength, frequency, and wave speed.

Wavelength ( $\lambda$ ): We use $k = 2\pi/\lambda$ . $\lambda = 2\pi/k = 2\pi/(2\pi) = 1$ m.
Frequency ( $f$ ): We use $\omega = 2\pi f$ . $f = \omega/(2\pi) = 4\pi/(2\pi) = 2$ Hz.
Wave Speed ( $v$ ): We can use either $v = \omega/k$ or $v=f\lambda$ . $v = (4\pi)/(2\pi) = 2$ m/s. Alternatively, $v = (2 \text{ Hz})(1 \text{ m}) = 2$ m/s.

Finally, what is the phase difference between two points separated by $0.25$ m? The phase difference, $\Delta\phi$ , depends only on the separation in space, $\Delta x$ .

\Delta\phi = k \cdot \Delta x = (2\pi) \cdot (0.25) = \frac{\pi}{2} \text{ radians}

This means the two points are a quarter of a cycle out of sync. When one is at a crest, the other is at the equilibrium position.

Quiz Questions 1/6

What fundamental characteristic distinguishes a progressive wave from Simple Harmonic Motion (SHM)?

Quiz Questions 2/6

In the wave function $y(x, t) = A \sin(kx - \omega t)$ , what does the angular wave number, $k$ , represent?

By understanding the structure of the wave function, you can deconstruct any wave's equation to reveal its physical properties.

Mastering CBSE Class 11 Physics Waves