explain and use the principle of superposition

Stationary (Standing) Waves – Principle of Superposition

Objective

Explain the principle of superposition and use it to derive the properties of stationary (standing) waves in strings and air columns. Apply these ideas to experimental set‑ups, energy considerations and typical Cambridge AS&A‑Level (9702) exam questions.

Key Concepts

• Superposition principle: When two or more waves occupy the same region of space, the resultant displacement at any point is the algebraic (vector) sum of the individual displacements.
• Interference: Constructive → larger amplitude; destructive → reduced or zero amplitude.
• Stationary (standing) wave: The pattern produced by the superposition of two identical waves travelling in opposite directions.
• Node: Point of permanently zero displacement (and zero velocity).
• Antinode: Point of maximum displacement (amplitude = ±2A for equal‑amplitude travelling waves).
• λ/2 rule: The distance between successive nodes (or successive antinodes) is always ½ λ.
• Boundary conditions: Physical constraints at the ends of the medium (fixed, free, open, closed) that determine which wavelengths are allowed.
• Energy distribution: In a standing wave kinetic and potential energy are stored in different parts of the pattern but the total energy is constant.

1. The Superposition Principle (General Form)

For any number of waves \(y_i(x,t)\) (\(i=1,2,\dots ,N\)) the resultant displacement is

\[ y(x,t)=\sum_{i=1}^{N} y_i(x,t). \]

The principle holds for sinusoidal, triangular, pulse or any other shape, provided the medium remains linear (no large‑amplitude non‑linear effects).

2. Formation of a Standing Wave

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

\[ y_1(x,t)=A\sin(kx-\omega t),\qquad y_2(x,t)=A\sin(kx+\omega t). \]

Applying superposition:

\[ \begin{aligned} y(x,t)&=y_1+y_2\\ &=2A\sin(kx)\cos(\omega t)\\ &=2A\cos(kx)\sin(\omega t)\quad\text{(using }\sin\alpha\cos\beta=\tfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\text{)}. \end{aligned} \]

Both forms represent a standing wave: the spatial factor (\(\sin kx\) or \(\cos kx\)) fixes the pattern of nodes and antinodes, while the temporal factor (\(\cos\omega t\) or \(\sin\omega t\)) makes every point oscillate with the same frequency \(\omega\).

3. Nodes, Antinodes and the λ/2 Rule

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

• Nodes: \(\sin(kx)=0\;\Rightarrow\;kx=n\pi\;(n=0,1,2,\dots)\). Hence \(x=n\frac{\lambda}{2}\).
• Antinodes: \(\sin(kx)=\pm1\;\Rightarrow\;kx=(n+\tfrac12)\pi\). Hence \(x=(n+\tfrac12)\frac{\lambda}{2}\).

Therefore the distance between two successive nodes (or two successive antinodes) is always \(\displaystyle\frac{\lambda}{2}\).

4. Boundary Conditions and Allowed Wavelengths

System	Boundary conditions	Allowed wavelengths \(\lambda_n\)	Fundamental frequency \(f_1\)
String	Both ends fixed	\(\displaystyle\lambda_n=\frac{2L}{n}\;(n=1,2,3,\dots)\)	\(\displaystyle f_1=\frac{v}{2L}\)
Open–Closed pipe (organ pipe)	One end open, one end closed	\(\displaystyle\lambda_n=\frac{4L}{2n-1}\;(n=1,2,3,\dots)\)	\(\displaystyle f_1=\frac{v}{4L}\)
Open–Open pipe (organ pipe)	Both ends open	\(\displaystyle\lambda_n=\frac{2L}{n}\;(n=1,2,3,\dots)\)	\(\displaystyle f_1=\frac{v}{2L}\)

5. Derivation of the Fundamental Frequency for a Fixed–Fixed String

1. For a string of length \(L\) with both ends fixed, the distance between adjacent nodes is \(\lambda/2\).
2. In the fundamental mode exactly one half‑wave fits into the string, so \(\lambda_1=2L\).
3. The wave speed on a stretched 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 \[ f_1=\frac{v}{\lambda_1}= \frac{1}{2L}\sqrt{\frac{T}{\mu}}. \]

6. Energy in a Standing Wave

For a sinusoidal standing wave \(y=2A\sin(kx)\cos(\omega t)\) on a string:

• Kinetic‑energy density (per unit length): \[ u_K=\frac12\mu\left(\frac{\partial y}{\partial t}\right)^2 =\frac12\mu\,(2A\omega)^2\sin^2(kx)\sin^2(\omega t). \] It is maximum where \(\sin(kx)=\pm1\) (the antinodes) and zero at the nodes.
• Potential‑energy density (due to tension): \[ u_P=\frac12T\left(\frac{\partial y}{\partial x}\right)^2 =\frac12T\,(2Ak)^2\cos^2(kx)\cos^2(\omega t). \] It is maximum at the nodes (\(\cos(kx)=\pm1\)) and zero at the antinodes.
• Total energy density: \[ u=u_K+u_P=\frac12\mu(2A\omega)^2\sin^2(kx)\sin^2(\omega t) +\frac12T(2Ak)^2\cos^2(kx)\cos^2(\omega t), \] which, after averaging over a full period, becomes spatially uniform. Hence the **total energy of a standing wave is constant**; energy is merely exchanged locally between kinetic and potential forms.

7. Worked Example – Fixed–Fixed String

Problem: A string 1.20 m long is fixed at both ends, under a tension of 80 N and with \(\mu=2.0\times10^{-3}\,\text{kg m}^{-1}\). Find the frequencies of the first three harmonics.

Solution:

1. Wave speed: \[ v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{80}{2.0\times10^{-3}}}=200\ \text{m s}^{-1}. \]
2. Fundamental frequency: \[ f_1=\frac{v}{2L}= \frac{200}{2\times1.20}=83.3\ \text{Hz}. \]
3. Higher harmonics are integer multiples of \(f_1\): \[ f_2=2f_1=166.7\ \text{Hz},\qquad f_3=3f_1=250.0\ \text{Hz}. \]

8. Worked Example – Open–Closed Pipe (Organ Pipe)

Problem: An open–closed pipe of length \(L=0.85\) m resonates at a frequency of \(f=210\) Hz. The speed of sound in air is \(v=340\) m s\(^{-1}\). Determine the harmonic number and the positions of the first two nodes measured from the closed end.

Solution:

1. Observed wavelength: \[ \lambda_{\text{obs}}=\frac{v}{f}= \frac{340}{210}=1.619\ \text{m}. \]
2. For an open–closed pipe \(\lambda_n=4L/(2n-1)\). Solving \(\lambda_{\text{obs}}=4L/(2n-1)\) gives \[ 2n-1=\frac{4L}{\lambda_{\text{obs}}}= \frac{4(0.85)}{1.619}\approx2.1\approx2, \] so \(2n-1=3\) and \(n=2\). The pipe is vibrating in the **second odd harmonic** (the 3rd harmonic overall for this type of pipe).
3. Wavelength for this mode: \[ \lambda_2=\frac{4L}{3}=1.133\ \text{m}. \] Distance between successive nodes is \(\lambda_2/2=0.567\) m. Nodes measured from the closed end:
   • Node 1 (closed end): \(x=0\) m
   • Node 2: \(x=0.567\) m
   The next node would lie at \(x=1.134\) m, outside the pipe, so only two nodes are present inside the pipe.

9. Experimental Techniques

• String experiment (rubber‑band or steel string on a speaker): One end of a taut string is attached to a loudspeaker that drives it sinusoidally. By varying the driving frequency, resonant patterns appear. Nodes are identified as points that remain still; antinodes vibrate with maximum amplitude. Measuring the distance between consecutive nodes gives \(\lambda/2\), from which the wave speed \(v\) can be calculated using \(v=f\lambda\).
• Air‑column (resonance tube) experiment: A tube closed at one end and open at the other contains a movable piston. A tuning fork of known frequency excites the column; the piston is adjusted until a loud resonance is heard. The piston marks a node; the open end is an antinode. The measured distance between node and antinode equals \(\lambda/4\). Example: if the piston is at 0.21 m from the closed end, \(\lambda=4\times0.21=0.84\) m and the speed of sound follows from \(v=f\lambda\).
• Microwave standing‑wave experiment: A microwave source and a flat metal reflector form a standing wave in free space. A small diode detector attached to a movable arm records intensity. Successive intensity minima (or maxima) are spaced by \(\lambda/2\).
  Formula: \(\displaystyle \lambda = 2\Delta x\), where \(\Delta x\) is the measured distance between two adjacent minima.
  Example: If the minima are 2.5 cm apart, \(\lambda = 2(2.5\ \text{cm}) = 5.0\ \text{cm}\). The corresponding frequency is \(f=v/\lambda\) (using \(v\approx3.00\times10^{8}\) m s\(^{-1}\) for microwaves).

10. Common Misconceptions

• Thinking a standing wave “travels” along the medium – the pattern is stationary; only the particles oscillate in place.
• Confusing a node (zero displacement) with a point of zero velocity – at a node the displacement is always zero, so the velocity is also zero at all times.
• Assuming any wavelength can exist in a bounded system – only wavelengths that satisfy the specific boundary conditions are permitted, giving a discrete set of resonant frequencies.
• Believing the distance between a node and the nearest antinode is \(\lambda\) – it is \(\lambda/4\) for a pipe with one end closed (or open) and \(\lambda/2\) for a string with both ends fixed.
• Neglecting the energy picture – kinetic energy is maximal at antinodes, potential energy at nodes; the total energy of a standing wave remains constant.

11. Summary

The principle of superposition allows two identical travelling waves moving in opposite directions to be added, producing a standing wave characterised by nodes, antinodes and a spatial spacing of \(\lambda/2\). Boundary conditions at the ends of a string or air column restrict the allowed wavelengths to discrete values, leading to a series of harmonics with frequencies \(f_n=nf_1\) (or \(f_n=(2n-1)f_1\) for open‑closed pipes). Energy in a standing wave is continuously exchanged between kinetic and potential forms, yet the total energy is constant. Mastery of these ideas enables you to analyse musical instruments, resonant air columns, microwave cavities and to answer the majority of Cambridge 9702 exam questions on Topic 8.