C.4.3—Strings and pipes
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Start with the boundary conditions
A fixed end of a string is a displacement node; a free end is a displacement antinode. For air displacement in a pipe, a closed end is a displacement node and an open end is a displacement antinode. These end conditions determine which standing-wave patterns are allowed.
Use the string patterns
For a string fixed at both ends, or with two free ends, the nth harmonic has n half-wavelengths in length L: λn=2L/n and fn=nv/(2L). For one fixed and one free end, the allowed patterns contain an odd number of quarter-wavelengths: λn=4L/(2n−1) and fn=(2n−1)v/(4L), with n=1,2,3,….
Apply the same geometry to pipes
An open pipe has displacement antinodes at both ends and follows the two-open-end pattern. A closed pipe has a displacement node at the closed end and an antinode at the open end, so only the odd sequence of harmonics is allowed. Use v=fλ after finding the wavelength from the boundary pattern. End corrections for open pipes are not required.
Common trap
Do not use the closed-pipe formula for an open pipe, and do not count pressure nodes or pressure antinodes here: the syllabus asks for air-displacement nodes and antinodes. Also use “first harmonic” for the lowest-frequency mode; the syllabus does not require the terms fundamental or overtone.
Questions ask for the wavelength or frequency sequence in strings or open/closed pipes. The decisive step is identifying the end conditions before applying a formula.
What expression / What is
Translate each end into a displacement node or antinode, fit the correct number of half- or quarter-wavelengths into L, then use v=fλ.
Applying f=nv/(2L) to a one-open-one-closed pipe, or counting pressure rather than air-displacement boundary conditions.
Representative question
Deduce that the length of the horn is about 0.20 m .
f1=4Lv,f2=3f1=4L3v;
C.4 is secure when you can move from boundary conditions and superposition to the observed response.