understand how wavelength may be determined from the positions of nodes or antinodes of a stationary wave

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

SystemEnd condition (displacement)Allowed wavelengthsFundamental \$\lambda_1\$
String (or air column) fixed at both endsNode – 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 displacementSame 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\$.

  • For a string under tension \$T\$ with linear mass density \$\mu\$: \$v=\sqrt{T/\mu}\$.

  • For sound in air at temperature \$T{\rm C}\$: \$v\approx 331+0.6\,T{\rm C}\;\text{m s}^{-1}\$.

  • Combining with the harmonic relations gives the resonant frequencies:

    \$\$f_n=\frac{nv}{2L}\;(N\!-\!N),\qquad

    f_n=\frac{(2n-1)v}{4L}\;(N\!-\!A).\$\$

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.

  1. Two node‑to‑node intervals \$\Rightarrow d_{NN}=0.45/2=0.225\,\$m.

  2. \$\lambda=2d_{NN}=0.45\,\$m.

  3. 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.

  4. 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.

  1. \$\lambda=2d_{AA}=0.60\,\$m.

  2. Fundamental condition \$L=\lambda/2\Rightarrow\lambda_{\rm theo}=2L=2.40\,\$m.

  3. 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.

  1. Effective length \$L_{\rm eff}=L+0.6r=0.80+0.006=0.806\,\$m.

  2. From \$d{NN}\$, \$\lambda=2d{NN}=0.36\,\$m.

  3. 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

ConfigurationMeasured distanceRelation 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)

  1. 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.

  2. Open–open pipe, \$L=1.20\,\$m: \$d_{AA}=0.30\,\$m.

    Answer: \$\lambda=2d_{AA}=0.60\,\$m → \$n=4\$ (fourth harmonic).

  3. 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).

  4. 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.

  5. 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.