explain and use the principle of superposition

Stationary Waves

Objective

Explain the principle of superposition and use it to derive the properties of stationary (standing) waves in strings and air columns.

Key Concepts

• Superposition principle: When two or more waves occupy the same region of space, the resultant displacement is the algebraic sum of the individual displacements.
• Interference: Constructive interference occurs when the displacements add to give a larger amplitude; destructive interference occurs when they cancel.
• Stationary (standing) wave: A pattern formed by the superposition of two identical waves travelling in opposite directions.
• Node: A point of zero displacement in a standing wave.
• Antinode: A point of maximum displacement in a standing wave.
• Boundary conditions: The constraints at the ends of the medium (fixed, free, open, closed) that determine the allowed wavelengths.

Mathematical Description

Consider two sinusoidal travelling waves of equal amplitude $A$, angular frequency $\omega$ and wave‑number $k$, moving in opposite directions along the $x$‑axis:

$$y_1(x,t)=A\sin(kx-\omega t)$$ $$y_2(x,t)=A\sin(kx+\omega t)$$

Applying the superposition principle, the resultant displacement is

$$y(x,t)=y_1+y_2=2A\sin(kx)\cos(\omega t)$$

or, using the identity $\sin\alpha\cos\beta = \tfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]$, the same result can be written as

$$y(x,t)=2A\cos(kx)\sin(\omega t)$$

Both forms describe a standing wave: the spatial part ($\sin(kx)$ or $\cos(kx)$) determines the positions of nodes and antinodes, while the temporal part ($\cos(\omega t)$ or $\sin(\omega t)$) describes the oscillation of each point.

Conditions for Nodes and Antinodes

From $y(x,t)=2A\sin(kx)\cos(\omega t)$:

• Nodes occur where $\sin(kx)=0 \;\Rightarrow\; kx=n\pi \;(n=0,1,2,\dots)$.
• Antinodes occur where $\sin(kx)=\pm1 \;\Rightarrow\; kx=(n+\tfrac12)\pi$.

Since $k=2\pi/\lambda$, the distance between adjacent nodes (or antinodes) is $\lambda/2$.

Allowed Wavelengths for Common Boundaries

System	Boundary Conditions	Allowed Wavelengths	Fundamental Frequency $f_1$
String	Both ends fixed	$\lambda_n = \dfrac{2L}{n}\;(n=1,2,3,\dots)$	$f_1 = \dfrac{v}{2L}$
Open–Closed Pipe (organ pipe)	One end open, one end closed	$\lambda_n = \dfrac{4L}{2n-1}\;(n=1,2,3,\dots)$	$f_1 = \dfrac{v}{4L}$
Open–Open Pipe (organ pipe)	Both ends open	$\lambda_n = \dfrac{2L}{n}\;(n=1,2,3,\dots)$	$f_1 = \dfrac{v}{2L}$

Derivation of the Fundamental Frequency for a Fixed–Fixed String

1. For a string of length $L$ with both ends fixed, the distance between two successive nodes is $\lambda/2$.
2. Therefore the fundamental mode (first harmonic) fits exactly half a wavelength in the string: $\lambda_1 = 2L$.
3. The wave speed on the string is $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ the linear mass density.
4. Using $v = f\lambda$, the fundamental frequency is $$f_1 = \frac{v}{\lambda_1}= \frac{1}{2L}\sqrt{\frac{T}{\mu}}.$$

Example Problem

Problem: A string 1.20 m long is fixed at both ends and under a tension of 80 N. Its linear mass density is $2.0\times10^{-3}\,\text{kg m}^{-1}$. Find the frequencies of the first three harmonics.

Solution:

1. Calculate the wave speed: $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{2.0\times10^{-3}}}= \sqrt{4.0\times10^{4}} = 200\ \text{m s}^{-1}.$$
2. Fundamental frequency: $$f_1 = \frac{v}{2L}= \frac{200}{2\times1.20}= \frac{200}{2.40}=83.3\ \text{Hz}.$$
3. Higher harmonics are integer multiples: $$f_2 = 2f_1 = 166.7\ \text{Hz},\qquad f_3 = 3f_1 = 250.0\ \text{Hz}.$$

Common Misconceptions

• Thinking that a standing wave “moves” along the medium – the pattern is stationary; only the particles oscillate.
• Confusing nodes with points of zero velocity – at a node the displacement is always zero, but the velocity is also zero at all times.
• Assuming any wavelength can exist in a bounded system – only those that satisfy the boundary conditions are allowed.

Summary

The principle of superposition allows us to combine two identical travelling waves moving in opposite directions to form a stationary wave. The resulting pattern is characterised by nodes and antinodes, with a spatial separation of $\lambda/2$. Boundary conditions dictate the permissible wavelengths and thus the discrete set of resonant frequencies (harmonics). Mastery of these concepts enables analysis of musical instruments, resonant cavities, and many A‑Level exam questions.