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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 4: Waves at Boundaries</b><br>
Thu 09 Oct 2025 13:00
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== Recap ==

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

And we also have: <div class="pn-import-equation"><nowiki>$$v = \frac{\omega}{k}$$</nowiki></div>

The former is useful explictly for a wave travelling over a string, while the latter is applicable to the movement of any wave. On a string, a higher tension yields a higher restoring force and therefore a higher speed, while a higher mass per unit length gives a higher mass for some arbitrary length of string, therefore a lower acceleration and lower speed.

== A Quick Interlude ==

The speed of a mechanical wave has the general form: <div class="pn-import-equation"><nowiki>$$v = \sqrt{\frac{\text{Restoring force returning to equilibrium}}{\text{Inertia resisting return to equilibrium}}}$$</nowiki></div>

== Reflection ==

When a wave hits a fixed boundary, it is reflected and inverted. Lets consider a case where a string is fixed on the LHS and is reflected back:

[[File:OW-Y1S1-lec04-02.png|thumb|center|600px]]

[[File:OW-Y1S1-lec04-03.png|thumb|center|600px|Or with two strings, going from thick to thin: ]]

[[File:OW-Y1S1-lec04-04.png|thumb|center|600px|Or from thin to thick.]]

== Waves Interacting at Boundaries ==

Say we have two pieces of string connected to each other, one thin string with mass per unit length <nowiki>$\mu_1$</nowiki> and a thicker string with <nowiki>$\mu_2$</nowiki> (both under the same tension, <nowiki>$T$</nowiki>). If a wave pulse is passed along from the thin string to the thicker string, whath happens at the point of connection, <nowiki>$P$</nowiki>?

[[File:OW-Y1S1-lec04-01.png|thumb|center|600px|The two connected strings.]]

Consider a travelling wave coming from the left: <div class="pn-import-equation"><nowiki>$$y_1 = A \cos(k_1 x - \omega_1 t)$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$y_2 = B \cos(k_2 x - \omega_2 t)$$</nowiki></div>

At the point of connection, we’ll define this as <nowiki>$x = 0$</nowiki>. Here (as the string is not broken, so it must be connected): <div class="pn-import-equation"><nowiki>\begin{equation}y_1 = y_2\end{equation}</nowiki></div>

Also, force is finite, therefore curvature must be finite (as force is proportional to curvature). Therefore we cannot have any discontinuities in curvature. <div class="pn-import-equation"><nowiki>\begin{equation}\frac{\partial y_1}{\partial x_1} = \frac{\partial y_2}{\partial x_2}\end{equation}</nowiki></div>

Disregarding non-linear effects, so assuming that the frequency with which the waves travel down the string is the same for both parts of the string, <nowiki>$\omega_1 = \omega_2 = \omega$</nowiki>.

From (1) and noting <nowiki>$x = 0$</nowiki>: <div class="pn-import-equation"><nowiki>\begin{equation}A \cos(-\omega t) = B \cos(\omega t)\end{equation}</nowiki></div>

And from (2) with the same note: <div class="pn-import-equation"><nowiki>\begin{equation}-k_1 A \sin -\omega t = -k_2 B \sin -\omega t\end{equation}</nowiki></div>

However this suggests that <nowiki>$A = B = 0$</nowiki>. This is technically a solution, but not really - it doesn’t represent an actual wave (just two flat lines with no amplitude ever)

=== Including Reflection ===

The previous case did not work as we disregarded reflection at the wave boundary, <nowiki>$P$</nowiki>. Lets add an extra wave (the C term) to represent the reflection back into the light string: <div class="pn-import-equation"><nowiki>$$y_1 = A \cos(k_1 x - \omega t) + C \cos(k_1 x + \omega t) = B \cos(k_2 x - \omega t)$$</nowiki></div>

From y1 = y2 at x = 0 we get: <div class="pn-import-equation"><nowiki>$$y_1 = A \cos(- \omega t) + C \cos( -\omega t) = B \cos( - \omega t)$$</nowiki></div> So: <nowiki>$A+C = B$</nowiki>

From (2) we get: <div class="pn-import-equation"><nowiki>$$-k_1 A \sin(- \omega t) - k_1 C \sin(\omega t) = -k_2 B \sin(-\omega t)$$</nowiki></div>

Which has solutions: <div class="pn-import-equation"><nowiki>$$B = \frac{2k_1}{k_1 + k_2} A$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$C = \frac{k_1 - k_2}{k_1 + k_2}A$$</nowiki></div>

We have the incident wave: <div class="pn-import-equation"><nowiki>$$y_1 = Acos(k_1 x - \omega t)$$</nowiki></div>

The transmitted wave: <div class="pn-import-equation"><nowiki>$$y_2 = B \cos(k_2 x - \omega t) = \frac{2k_1}{k_1 + k_2} A \cos(k_2 x - \omega t)$$</nowiki></div>

And lastly the reflected wave: <div class="pn-import-equation"><nowiki>$$y_3 = C \cos(k_1 x + \omega t) = \frac{k_1 - k_2}{k_1 + k_2} A \cos(k_1 x - \omega t)$$</nowiki></div>

We note:

<nowiki>$k_1$</nowiki> is the wave number in the medium where the incident wave comes from.

<nowiki>$k_2$</nowiki> is the wave number in the medium where the transmitted wave goes into.

=== Example ===

If the wave comes in from the left, then <nowiki>$\mu_2 > \mu_1$</nowiki> per example diagram, then <nowiki>$k_2 > k_1$</nowiki>, and <nowiki>$C$</nowiki> is negative.

Note that if the single pulse as a postive y amplitude, then the transmitted pulse in the heavier string will also have a positive y amplitude, but will have a smaller magnitude. The reflected wave will have a negative y-amplitude as the reflection inverts it.

If the wave comes from the right (from thick to thin), then <nowiki>$\mu_1 > \mu_2$</nowiki>, <nowiki>$k_1 > k_2$</nowiki> and C is positive.

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