Stationary (Standing) Waves – Cambridge IGCSE / A‑Level Physics 9702
1. Superposition – how a standing wave is formed
- Two identical travelling waves moving in opposite directions:
\$y1(x,t)=A\sin(kx-\omega t),\qquad y2(x,t)=A\sin(kx+\omega t)\$
- Using the identity \$\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\$ gives
\$y(x,t)=y1+y2=2A\cos(kx)\sin(\omega t)\$
- The spatial factor \$\cos(kx)\$ is independent of time – points where \$\cos(kx)=0\$ never move (nodes); points where \$|\cos(kx)|=1\$ have maximum amplitude (antinodes).
- Key relation: \$k=2\pi/\lambda,\;\;\omega=2\pi f\$.
2. Nodes, antinodes and energy distribution
- Node (N): \$y=0\$ for all \$t\$. Distance between successive nodes \$d_{NN}= \lambda/2\$.
- Antinode (A): \$|y|=2A\$ (maximum displacement). Distance between successive antinodes \$d_{AA}= \lambda/2\$.
- Distance from a node to the nearest antinode \$d_{NA}= \lambda/4\$.
- Energy: kinetic energy is zero at nodes, potential energy is zero at antinodes; the total energy is constant and stored alternately as kinetic and potential along the wave.
3. Boundary conditions, harmonic series & real‑world corrections
| System | End condition (displacement) | Allowed wavelengths | Fundamental \$\lambda_1\$ |
|---|
| String (or air column) fixed at both ends | Node – Node (N–N) | \$\displaystyle \lambda_n=\frac{2L}{n}\qquad n=1,2,3,\dots\$ | \$\lambda_1=2L\$ |
| Open–open pipe (or string with both ends free) | Antinode – Antinode (A–A) | \$\displaystyle \lambda_n=\frac{2L}{n}\qquad n=1,2,3,\dots\$ | \$\lambda_1=2L\$ |
| Open–closed pipe (or string fixed at one end, free at the other) | Node – Antinode (N–A) | \$\displaystyle \lambda_n=\frac{4L}{2n-1}\qquad n=1,2,3,\dots\$ | \$\lambda_1=4L\$ |
| Closed–closed pipe (pressure nodes) | Node – Node (N–N) for pressure, Antinode – Antinode for displacement | Same as fixed–fixed: \$\lambda_n=2L/n\$ | \$\lambda_1=2L\$ |
End‑corrections (practical note): For air columns the effective length is \$L{\rm eff}=L+0.6r\$ (open end) or \$L{\rm eff}=L+0.3r\$ (closed end), where \$r\$ is the pipe radius. Include this when comparing measured and theoretical \$\lambda\$.
4. Determining wavelength from node/antinode positions
From the geometry in section 2:
- Successive nodes (or antinodes): \$d{NN}=d{AA}= \lambda/2\;\Rightarrow\;\lambda = 2d\$.
- Node to the nearest antinode: \$d_{NA}= \lambda/4\;\Rightarrow\;\lambda = 4d\$.
If \$N\$ equal node‑to‑node intervals are measured over a length \$L_{\rm meas}\$,
\$L{\rm meas}=N\frac{\lambda}{2}\;\Longrightarrow\;\lambda=\frac{2L{\rm meas}}{N}.\$
Uncertainty propagation (AO3) – for \$\lambda =2d\$,
\$\Delta\lambda =2\,\Delta d,\$
and for \$\lambda =\dfrac{2L_{\rm meas}}{N}\$,
\$\Delta\lambda =\frac{2}{N}\,\Delta L_{\rm meas}.\$
Include systematic uncertainties (e.g. end‑correction, ruler calibration) in the final error budget.
5. Linking wavelength, frequency and wave speed
For any wave \$v = f\lambda\$.
6. Practical skills required for Paper 3/5 (AO3)
- Planning the set‑up: choose a suitable driver (tuning fork, speaker, electromagnetic shaker), secure the medium, and decide how nodes/antinodes will be visualised (dust, laser pointer, oscilloscope).
- Data collection: record several consecutive node‑to‑node (or node‑to‑antinode) distances, repeat the measurement for at least two harmonics, and note the driving frequency (from a calibrated source).
- Uncertainty analysis: estimate random error (standard deviation of repeated \$d\$ values) and systematic error (ruler resolution, end‑correction). Propagate to \$\lambda\$ and, if required, to \$v\$ or \$f\$.
- Evaluation: compare experimental \$\lambda\$ with the theoretical value from the harmonic formula, discuss possible reasons for discrepancy (energy loss, non‑ideal boundary, temperature variation).
7. Diffraction – single‑slit
- Condition for minima: \$a\sin\theta = m\lambda\;(m=1,2,3,\dots)\$ where \$a\$ is slit width.
- For the first minimum (\$m=1\$) the angular width \$\Delta\theta\approx\lambda/a\$.
Worked example: A laser (\$\lambda=632.8\,\$nm) passes through a slit \$a=0.10\,\$mm. Find the angle to the first dark fringe.
\$\sin\theta=\frac{\lambda}{a}= \frac{6.328\times10^{-7}}{1.0\times10^{-4}}=6.33\times10^{-3}\;\Rightarrow\;\theta\approx0.36^{\circ}.\$
8. Interference – double‑slit
- Path‑difference condition for bright fringes: \$d\sin\theta = n\lambda\;(n=0,\pm1,\pm2,\dots)\$ where \$d\$ is slit separation.
- Fringe spacing on a screen at distance \$D\$: \$y_n = \dfrac{n\lambda D}{d}\$ (small‑angle approximation).
Worked example: \$d=0.30\,\$mm, \$D=2.00\,\$m, \$\lambda=500\,\$nm. Find the separation of the \$n=2\$ bright fringe from the central maximum.
\$y_2=\frac{2\lambda D}{d}= \frac{2(5.0\times10^{-7})(2.0)}{3.0\times10^{-4}}=6.7\times10^{-3}\,\text{m}=6.7\,\$mm.
9. Diffraction grating
- Grating equation: \$d\sin\theta = n\lambda\$, where \$d\$ is the grating spacing (inverse of lines per metre).
- Higher orders (\$n=2,3,\dots\$) are observable when \$\sin\theta\le1\$.
Worked example: A grating has \$600\$ lines cm\$^{-1}\$. Light of unknown wavelength produces a first‑order maximum at \$\theta=20^{\circ}\$. Find \$\lambda\$.
Grating spacing \$d=1/(600\times10^{2})=1.67\times10^{-6}\,\$m.
\$\lambda = d\sin\theta = 1.67\times10^{-6}\sin20^{\circ}=5.7\times10^{-7}\,\text{m}=570\,\$nm.
10. Worked examples for standing waves
Example 1 – Fixed–fixed string, second harmonic
Length \$L=1.20\,\$m. Measured distance between the first and third nodes \$=0.45\,\$m.
- Two node‑to‑node intervals \$\Rightarrow d_{NN}=0.45/2=0.225\,\$m.
- \$\lambda=2d_{NN}=0.45\,\$m.
- For \$n=2\$, theoretical \$\lambda=2L/n=2(1.20)/2=1.20\,\$m. The measured \$0.45\,\$m corresponds to half a wavelength of the second harmonic, confirming the identification.
- Uncertainty (ruler \$\pm0.5\,\$mm): \$\Delta\lambda=2\Delta d=1.0\,\$mm.
Example 2 – Open–open pipe, fundamental
Pipe length \$L=1.20\,\$m. Distance between two consecutive antinodes \$d_{AA}=0.30\,\$m.
- \$\lambda=2d_{AA}=0.60\,\$m.
- Fundamental condition \$L=\lambda/2\Rightarrow\lambda_{\rm theo}=2L=2.40\,\$m.
- Measured \$\lambda\$ is a quarter of the fundamental, so the pipe is resonating in the fourth harmonic (\$n=4\$): \$L=n\lambda/2=4(0.60)/2=1.20\,\$m.
Example 3 – Closed–closed tube with end‑correction
Physical length \$L=0.80\,\$m, radius \$r=0.010\,\$m. Measured node‑to‑node distance \$d_{NN}=0.18\,\$m.
- Effective length \$L_{\rm eff}=L+0.6r=0.80+0.006=0.806\,\$m.
- From \$d{NN}\$, \$\lambda=2d{NN}=0.36\,\$m.
- Fundamental for a closed–closed tube: \$L{\rm eff}=\lambda/2\Rightarrow\lambda{\rm theo}=2L{\rm eff}=1.612\,\$m. The measured value therefore belongs to the third harmonic (\$n=3\$): \$\lambda=2L{\rm eff}/3=1.612/3=0.537\,\$m, consistent with the experimental estimate within uncertainty.
11. Quick‑reference summary table
| Configuration | Measured distance | Relation to \$\lambda\$ | Formula for \$\lambda\$ |
|---|
| Successive nodes (or antinodes) | \$d{NN}=d{AA}\$ | \$d=\lambda/2\$ | \$\lambda=2d\$ |
| Node to adjacent antinode | \$d_{NA}\$ | \$d=\lambda/4\$ | \$\lambda=4d\$ |
| Fixed–fixed string (fundamental) | Length \$L\$ | \$L=\lambda/2\$ | \$\lambda=2L\$ |
| Open–closed pipe (fundamental) | Length \$L\$ | \$L=\lambda/4\$ | \$\lambda=4L\$ |
| Open–open or closed–closed pipe (fundamental) | Length \$L\$ | \$L=\lambda/2\$ | \$\lambda=2L\$ |
| Single‑slit diffraction (first minimum) | Slit width \$a\$, angle \$\theta\$ | \$a\sin\theta=\lambda\$ | \$\lambda=a\sin\theta\$ |
| Double‑slit interference (bright fringe \$n\$) | Separation \$d\$, screen distance \$D\$, fringe position \$y_n\$ | \$y_n=\dfrac{n\lambda D}{d}\$ | \$\lambda=\dfrac{y_n d}{n D}\$ |
| Diffraction grating (order \$n\$) | Grating spacing \$d\$, angle \$\theta\$ | \$d\sin\theta=n\lambda\$ | \$\lambda=\dfrac{d\sin\theta}{n}\$ |
12. Practice questions (with concise answer keys)
- String, second harmonic: distance between the first and third nodes \$=0.45\,\$m.
Answer: \$d{NN}=0.225\,\$m → \$\lambda=2d{NN}=0.45\,\$m.
- Open–open pipe, \$L=1.20\,\$m: \$d_{AA}=0.30\,\$m.
Answer: \$\lambda=2d_{AA}=0.60\,\$m → \$n=4\$ (fourth harmonic).
- Closed–closed tube: \$d{NN}=0.18\,\$m, \$d{NA}=0.18\,\$m.
Answer: \$\lambda\$ from \$d{NN}=0.36\,\$m, from \$d{NA}=0.72\,\$m → average \$0.54\,\$m (consistent with \$n=3\$ if \$L_{\rm eff}\approx0.81\,\$m).
- Single‑slit diffraction: \$a=0.20\,\$mm, first dark fringe at \$\theta=2.0^{\circ}\$. Find \$\lambda\$.
Answer: \$\lambda=a\sin\theta=2.0\times10^{-4}\times\sin2^{\circ}=6.98\times10^{-6}\,\$m \$=698\,\$nm.
- Double‑slit interference: \$d=0.40\,\$mm, \$D=1.5\,\$m, \$y_{3}=9.0\,\$mm. Find \$\lambda\$.
Answer: \$\lambda=\dfrac{y_3 d}{3D}= \dfrac{9.0\times10^{-3}\times4.0\times10^{-4}}{3\times1.5}=8.0\times10^{-7}\,\$m \$=800\,\$nm.
13. Suggested diagram (to be drawn by the student)
Sketch a standing wave on a string showing:
- Four consecutive nodes \$N1\$–\$N4\$.
- Three antinodes \$A1\$–\$A3\$ between them.
- Arrows indicating the measured distances \$d{NN}\$ (between \$N1\$–\$N2\$, etc.) and \$d{NA}\$ (between \$N1\$–\$A1\$, etc.).
- Labels stating \$d{NN}= \lambda/2\$ and \$d{NA}= \lambda/4\$.
This diagram is useful for answering “identify the node‑to‑node distance” and “show how you would calculate \$\lambda\$” in exam questions.