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Import converted Year 1 Semester 1 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<div class="pn-lecture-meta">
<b>Lecture 3: Generalised Wavefunctions</b><br>
Wed 08 Oct 2025 11:00
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[[Year 1 Sem 1/LC Optics and Waves/Lecture 02|← Previous lecture]] · [[Year 1 Sem 1/LC Optics and Waves|Module]] · [[Year 1 Sem 1/LC Optics and Waves/Lecture 04|Next lecture →]]

Recap: For a sine wave, the wavefunction is: <div class="pn-import-equation"><nowiki>$$y(x,t) = A \cos(kx - \omega t)$$</nowiki></div> Where k is the wave number, <nowiki>$k = \frac{2 \pi}{\lambda}$</nowiki> and omega is <nowiki>$\frac{2 \pi}{T} = 2 \pi f$</nowiki>

== More Wavefunctions ==

What is the general form of a triangular shaped wave? What about a square stepped wave? TODO: Diagram

We know that each particle (i.e. along a string) copies the motion of its immediate left-hand neighbour (for a particle moving in the positive x), with a time delay proportional to their distance. Every wave is describable <nowiki>$y(x,t)$</nowiki> wave function, and every mechanical wave relies on a medium (i.e. a piece of string or water, concrete etc) to travel.

Ideally, we want to be able to find a general form of a wave function.

In a single fixed instant of time, the moving wave pulse is stationary, so purely a function of <nowiki>$y = f(x)$</nowiki>. We want a wave function where we can input any value of <nowiki>$t$</nowiki>, so we need a moving frame of reference. We define this frame of reference as <nowiki>$O'$</nowiki> (for the origin) and <nowiki>$x', y'$</nowiki> axes. This frame of reference moves with the wave pulse and at the same speed, therefore y’ is a function of x’ only, independant of speed.

=== New System ===

<nowiki>$x$</nowiki> is the distance from the origin <nowiki>$O$</nowiki> to the relevant point, while <nowiki>$x'$</nowiki> is the distance from <nowiki>$O'$</nowiki>.

<div class="pn-import-equation"><nowiki>$$x = x' + vt$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$x' = x - vt$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$y' = f(x') = f(x-vt)$$</nowiki></div>

However, as the wave is moving purely in one direction (along x), <nowiki>$y = y'$</nowiki>, so: <div class="pn-import-equation"><nowiki>$$y = f(x - vt)$$</nowiki></div>

=== Back to Basics ===

Going back to: <div class="pn-import-equation"><nowiki>$$y(x,t) = A \cos(kx - \omega t)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= A \cos\left[k\left(x - \frac{\omega}{k}t\right)\right]$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= A \cos\left[k\left(x - vt\right)\right]$$</nowiki></div> (Note, this is true for a wave in the positive x, for a wave moving in the negative x this would be x + vt)

=== Equivalent Rrepresentations ===

There are some equivalent representations for a sine wave: <div class="pn-import-equation"><nowiki>$$y(x,t) = A \cos \left(2 \pi f \frac{x}{v} - \omega t\right) = A \cos\left(2 \pi \left[\frac{x}{\lambda} - \frac{t}{T}\right]\right) = TODO$$</nowiki></div>

== More Differentiation ==

We’ve already looked at differentiating with respect to t, but what about x? This would give us the slope of the string at that point:

<div class="pn-import-equation"><nowiki>$$\frac{\partial y(x, t)}{\partial x}$$</nowiki></div>

And the curvature of the string: <div class="pn-import-equation"><nowiki>$$\frac{\partial^2 y(x, t)}{\partial x^2} = -k^2 A \cos(kx - \omega t) = -k^2 y(x, t)$$</nowiki></div> Note the similarities here with the equation for transverse acceleration:

<div class="pn-import-equation"><nowiki>$$(1) \frac{\partial^2y(x, t)}{\partial t^2} = -\omega^2 y(x, t)$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$(2) \frac{\partial^2 y(x, t)}{\partial x^2} = -k^2 y(x, t)$$</nowiki></div>

Dividing (1) by (2):

<div class="pn-import-equation"><nowiki>$$\frac{\partial^2y(x, t) / \partial t^2}{\partial^2 y(x, t)/\partial x^2} = \frac{\omega^2}{k^2} = v^2$$</nowiki></div>

Therefore: <div class="pn-import-equation"><nowiki>$$\frac{\partial^2 y(x, t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y(x, t)}{\partial t^2}$$</nowiki></div> We call this the ‘wave equation’ as every wave function y(x, t) must satisfy it, regardless of whether or not it is periodic or its direction of travel. If y(x, t) does not satisfy this, it is not a wave function.

=== An Example ===

<div class="pn-import-equation"><nowiki>$$y(x, t) = \frac{x^3 - vt^2}{e^t}$$</nowiki></div> TODO, the example in our own time

== Wave Equation for a String ==

[[File:OW-Y1S1-lec03-02.png|thumb|center|600px|A snapshot of the wave.]]

Lets say we have some string, suspended horizontally under tension. We generate a single wave pulse and allow it to propogate down the string.

We assume the string is 1D, under tension <nowiki>$T$</nowiki> (constant throughout) and has mass per unit length <nowiki>$\mu$</nowiki>.

Consider a small segment of string of length <nowiki>$\Delta L$</nowiki> from <nowiki>$x$</nowiki> to <nowiki>$x + \Delta x$</nowiki>. This string makes angle <nowiki>$\theta$</nowiki> with the horizontal at the bottom of the string, and angle <nowiki>$\theta'$</nowiki> with the horizontal at the top of the string.

Net force (transverse in y) is: <div class="pn-import-equation"><nowiki>$$F_y = T \sin \theta' - T \sin \theta$$</nowiki></div>

And using the small angle approx <nowiki>$\sin \theta \approx \tan \theta = \frac{dy}{dx}$</nowiki>:

<div class="pn-import-equation"><nowiki>$$F_y = T\left(\frac{dy}{dx} \Bigr\rvert_{x + \Delta x} - \frac{dy}{dx} \Bigr\rvert_{x}\right)$$</nowiki></div>

Using differentiation by first principles: <div class="pn-import-equation"><nowiki>$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$</nowiki></div>

We can say: <div class="pn-import-equation"><nowiki>$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \lim_{\Delta x \to 0} \frac{\frac{dy}{dx} \Bigr\rvert_{x + \Delta x} - \frac{dy}{dx} \Bigr\rvert_{x}}{\Delta x}$$</nowiki></div>

Therefore: <div class="pn-import-equation"><nowiki>$$F_y = T \frac{d^2y}{dx} \Delta x$$</nowiki></div>

The mass of this section of string is <nowiki>$\mu \Delta x$</nowiki>, and considering the acceleration in the y direction we can plug into <nowiki>$F = ma$</nowiki> to get:

<div class="pn-import-equation"><nowiki>$$F = ma$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$T \frac{d^2y}{dx^2} \Delta x = \mu \Delta x \frac{d^2 y}{dt^2}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$\frac{d^2y}{dx^2} = \frac{\mu}{T} \frac{d^2y}{dt^2}$$</nowiki></div>

The LHS is the rate of change of the string’s gradient, which as mentioned is the curvature of the string. The RHS includes the transverse acceleration of the string, therefore acceleration is proportional to curvature. To evaluate this at a fixed time/position we should write: <div class="pn-import-equation"><nowiki>$$\frac{\partial^2 y}{\partial x^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2}$$</nowiki></div>

And comparing to the wave equation we get: <div class="pn-import-equation"><nowiki>$$\frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$\frac{1}{v^2} = \frac{\mu}{T}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$v = \sqrt{\frac{T}{\mu}}$$</nowiki></div>

So a wave travels faster under a higher tension with a lower mass per unit length.

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Ash: Import converted Year 1 Semester 1 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc