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.
For any number of waves \(y_i(x,t)\) (\(i=1,2,\dots ,N\)) the resultant displacement is
\[
y(x,t)=\sum{i=1}^{N} yi(x,t).
\]
The principle holds for sinusoidal, triangular, pulse or any other shape, provided the medium remains linear (no large‑amplitude non‑linear effects).
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:
\[
\begin{aligned}
y(x,t)&=y1+y2\\
&=2A\sin(kx)\cos(\omega t)\\
&=2A\cos(kx)\sin(\omega t)\quad\text{(using }\sin\alpha\cos\beta=\tfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\text{)}.
\end{aligned}
\]
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 \(\omega\).
From \(y=2A\sin(kx)\cos(\omega t)\):
Hence \(x=n\frac{\lambda}{2}\).
Hence \(x=(n+\tfrac12)\frac{\lambda}{2}\).
Therefore the distance between two successive nodes (or two successive antinodes) is always \(\displaystyle\frac{\lambda}{2}\).
| 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 | \(\displaystyle\lambda_n=\frac{2L}{n}\;(n=1,2,3,\dots)\) | \(\displaystyle f_1=\frac{v}{2L}\) |
\[
f1=\frac{v}{\lambda1}= \frac{1}{2L}\sqrt{\frac{T}{\mu}}.
\]
For a sinusoidal standing wave \(y=2A\sin(kx)\cos(\omega t)\) on a string:
\[
u_K=\frac12\mu\left(\frac{\partial y}{\partial t}\right)^2
=\frac12\mu\,(2A\omega)^2\sin^2(kx)\sin^2(\omega t).
\]
It is maximum where \(\sin(kx)=\pm1\) (the antinodes) and zero at the nodes.
\[
u_P=\frac12T\left(\frac{\partial y}{\partial x}\right)^2
=\frac12T\,(2Ak)^2\cos^2(kx)\cos^2(\omega t).
\]
It is maximum at the nodes (\(\cos(kx)=\pm1\)) and zero at the antinodes.
\[
u=uK+uP=\frac12\mu(2A\omega)^2\sin^2(kx)\sin^2(\omega t)
+\frac12T(2Ak)^2\cos^2(kx)\cos^2(\omega t),
\]
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.
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:
\[
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{80}{2.0\times10^{-3}}}=200\ \text{m s}^{-1}.
\]
\[
f_1=\frac{v}{2L}= \frac{200}{2\times1.20}=83.3\ \text{Hz}.
\]
\[
f2=2f1=166.7\ \text{Hz},\qquad
f3=3f1=250.0\ \text{Hz}.
\]
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:
\[
\lambda_{\text{obs}}=\frac{v}{f}= \frac{340}{210}=1.619\ \text{m}.
\]
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).
\[
\lambda_2=\frac{4L}{3}=1.133\ \text{m}.
\]
Distance between successive nodes is \(\lambda_2/2=0.567\) m.
Nodes measured from the closed end:
The next node would lie at \(x=1.134\) m, outside the pipe, so only two nodes are present inside the pipe.
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).
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 \(fn=nf1\) (or \(fn=(2n-1)f1\) 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.
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