Year 1 Sem 1/LC Optics and Waves/Lecture 04

Lecture 4: Waves at Boundaries
Thu 09 Oct 2025 13:00

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Recap

We previously derived:

$$v = \sqrt{\frac{T}{\mu}}$$

And we also have:

$$v = \frac{\omega}{k}$$

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:

$$v = \sqrt{\frac{\text{Restoring force returning to equilibrium}}{\text{Inertia resisting return to equilibrium}}}$$

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:

Or with two strings, going from thick to thin:

Waves Interacting at Boundaries

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

Consider a travelling wave coming from the left:

$$y_1 = A \cos(k_1 x - \omega_1 t)$$

$$y_2 = B \cos(k_2 x - \omega_2 t)$$

At the point of connection, we’ll define this as $x = 0$. Here (as the string is not broken, so it must be connected):

\begin{equation}y_1 = y_2\end{equation}

Also, force is finite, therefore curvature must be finite (as force is proportional to curvature). Therefore we cannot have any discontinuities in curvature.

\begin{equation}\frac{\partial y_1}{\partial x_1} = \frac{\partial y_2}{\partial x_2}\end{equation}

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, $\omega_1 = \omega_2 = \omega$.

From (1) and noting $x = 0$:

\begin{equation}A \cos(-\omega t) = B \cos(\omega t)\end{equation}

And from (2) with the same note:

\begin{equation}-k_1 A \sin -\omega t = -k_2 B \sin -\omega t\end{equation}

However this suggests that $A = B = 0$. 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, $P$. Lets add an extra wave (the C term) to represent the reflection back into the light string:

$$y_1 = A \cos(k_1 x - \omega t) + C \cos(k_1 x + \omega t) = B \cos(k_2 x - \omega t)$$

From y1 = y2 at x = 0 we get:

$$y_1 = A \cos(- \omega t) + C \cos( -\omega t) = B \cos( - \omega t)$$

So: $A+C = B$ From (2) we get:

$$-k_1 A \sin(- \omega t) - k_1 C \sin(\omega t) = -k_2 B \sin(-\omega t)$$

Which has solutions:

$$B = \frac{2k_1}{k_1 + k_2} A$$

$$C = \frac{k_1 - k_2}{k_1 + k_2}A$$

We have the incident wave:

$$y_1 = Acos(k_1 x - \omega t)$$

The transmitted wave:

$$y_2 = B \cos(k_2 x - \omega t) = \frac{2k_1}{k_1 + k_2} A \cos(k_2 x - \omega t)$$

And lastly the reflected wave:

$$y_3 = C \cos(k_1 x + \omega t) = \frac{k_1 - k_2}{k_1 + k_2} A \cos(k_1 x - \omega t)$$

We note:

$k_1$ is the wave number in the medium where the incident wave comes from.

$k_2$ is the wave number in the medium where the transmitted wave goes into.

Example

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

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