Stationary-Wave Experiments | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Stationary-Wave Experiments

The same method works for a stretched string, microwaves and an air column: create a reflection, locate neighbouring nodes or antinodes, then convert their spacing into wavelength.

End corrections are ignored at this syllabus level.

Three experiments, one measurement idea

Three hand-drawn stationary-wave experiments: a vibrating stretched string, microwaves between a transmitter and reflector, and a tuning fork above an adjustable air column — The same measurement idea in three media. Left: stretched string. Centre: microwaves reflected by a metal plate. Right: resonance in an air column.

Setup	How the reflection is made	What is detected
stretched string	wave reflects at the fixed end	points of zero and maximum string motion
microwaves	wave reflects from a metal plate	alternating detector minima and maxima
air column	sound reflects at the closed water surface or tube end	resonance: a large sound amplitude

Turn the measured spacing into wavelength

On a string or microwave track, measure across several adjacent nodes or antinodes and divide by the number of spaces. Then double the adjacent spacing to obtain $\lambda$ .
For a string fixed at both ends in its simplest mode, both ends are nodes and $L=\lambda/2$ .
For an air column closed at one end and open at the other in its simplest mode, the closed end is a displacement node, the open end is a displacement antinode, and $L=\lambda/4$ .

fundamental λ = 4l

Once the wavelength and source frequency are known, use $v=f\lambda$ to determine the wave speed.

Common mistake

Decide whether the diagram shows displacement or pressure. At a closed end, air-particle displacement has a node but pressure has an antinode.

Your turn9702/11/M/J/25 Q31

A pipe is closed at one end. Its fundamental frequency is 820 Hz and the speed of sound is 330 m s⁻¹. Calculate the pipe length. [2] (9702/11/M/J/25 Q31)

Show worked answer

$\lambda=v/f=330/820=0.402\ \text{m}$ .
For a closed pipe, $L=\lambda/4$ .

Answer

L=0.10\ \text{m}