A 2013 UNSW acoustics study measured a Yolŋu-crafted didgeridoo from Arnhem Land: total length ~1.5 m (closed-open pipe), giving a fundamental $f_1 = v/4L \approx 340/(4 \times 1.5) \approx 65$ Hz. The first overtone (3rd harmonic, first allowed above fundamental for a closed-open pipe) was measured at 195 Hz = 3 × 65 Hz. The even 2nd harmonic at 130 Hz was absent, confirming the odd-harmonics-only rule for closed-open pipes.
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
A pipe open at both ends has length $L$. Predict the longest wavelength that can form a standing wave inside it. Write your prediction.
Warm-up, at an open end of a pipe, the standing wave has a:
Core Content
The 2013 UNSW acoustics team placed a microphone at the bell of a Yolŋu didgeridoo (~1.5 m long) and read the frequency spectrum on a laptop. The display showed a tall spike at 65 Hz, nothing at 130 Hz, another spike at 195 Hz, nothing at 260 Hz, then a spike at 325 Hz. Every second harmonic was absent. The pattern, fundamental, skip, skip two, present, skip, present, is the acoustic fingerprint of a closed-open pipe: only odd harmonics allowed.
A string fixed at both ends has nodes at both ends. The condition $L = n\lambda/2$ must be satisfied, giving harmonics:
$f_n = \dfrac{nv}{2L}$ ($n = 1, 2, 3, \ldots$)
An open-open pipe has antinodes at both ends; the same formula applies because the boundary condition is symmetric. Harmonics: $f_1, 2f_1, 3f_1, \ldots$
For a string fixed at both ends (or an open-open pipe), $f_n = nv/(2L)$, all harmonics are present ($n = 1, 2, 3, \ldots$). Nodes form at both fixed/closed ends; antinodes form at both open ends. The fundamental ($n = 1$) has $\lambda_1 = 2L$.
Pause, copy the highlighted formula and boundary conditions into your book before the check below.
Drive the wave superposition lab — driving a fixed 2.00 m string at 2, 4, 6 and 8 Hz builds the first four harmonics, and the node spacing you can read off the ruler is half a wavelength every time. It is in Lesson 4 , in the step called The Superposition Principle .
We just saw that fixed strings and open-open pipes produce all harmonics via $f_n = nv/(2L)$. That raises a question: what happens when one end is closed, does the formula just change, or does it remove some harmonics entirely? This card answers it → closing one end imposes a node there, which forces $L = (2n-1)\lambda/4$ and eliminates every even harmonic.
A pipe with one closed end has a node at the closed end and an antinode at the open end. The condition is $L = (2n-1)\lambda/4$, giving only odd harmonics:
$f_n = \dfrac{(2n-1)v}{4L}$ ($n = 1, 2, 3, \ldots$)
$n=1$: fundamental; $n=2$: 3rd harmonic ($3f_1$); $n=3$: 5th harmonic ($5f_1$) …
For a closed-open pipe, $f_n = (2n-1)v/(4L)$, only odd harmonics (1st, 3rd, 5th, …). The closed end is a node; the open end is an antinode. The fundamental ($n = 1$) has $\lambda_1 = 4L$, twice as long as an open-open pipe of the same length.
Add the highlighted formula and odd-harmonics rule to your notes before moving on.
A closed-open pipe of length 0.85 m has $v_{sound} = 340$ m/s. The fundamental frequency is:
A closed-open pipe produces harmonics at $f_1, 2f_1, 3f_1, \ldots$
A fixed string of length $L$ vibrating in its fundamental mode has $\lambda = 2L$.
Activities
Use $v_{sound} = 340$ m/s for all parts:
A didgeridoo (closed-open pipe) is 1.3 m long. Calculate: (a) the fundamental frequency, (b) the frequency of the next allowed harmonic, (c) explain why the 2nd harmonic ($2f_1$) is not produced.
Draw and label the standing wave patterns for the first three modes of a closed-open pipe. For each mode, label: node at closed end, antinode at open end, and the number of quarter-wavelengths.
Which is the odd one out? Three of these are allowed harmonics in a closed-open pipe (fundamental $f_1$).
An open-open pipe 0.34 m long has $v = 340$ m/s. The frequency of the third harmonic is:
Doubling the length of a closed-open pipe while keeping wave speed constant:
UnderstandBand 3(3 marks) 1. Explain why a closed-open pipe produces only odd harmonics. Use the boundary conditions at each end in your answer.
ApplyBand 4(3 marks) 2. A flute (open-open pipe) has an effective length of 0.66 m. Calculate the frequencies of the first three harmonics ($v = 340$ m/s).
AnalyseBand 5(4 marks) 3. A musician shortens a clarinet (closed-open pipe) by half. Describe and explain the effect on (a) the fundamental frequency and (b) the set of allowed harmonics.
1. Open-open 0.5 m: $f_1 = 340/(2×0.5) = 340$ Hz; $f_2 = 680$ Hz; $f_3 = 1020$ Hz
2. Closed-open 0.5 m: $f_1 = 340/(4×0.5) = 170$ Hz; 3rd harmonic = $3 × 170 = 510$ Hz
3. Guitar string: $f_1 = 400/(2×0.65) = 307.7$ Hz
(a) $f_1 = 340/(4×1.3) = 65.4$ Hz (b) $3f_1 = 196$ Hz (c) The node-at-closed-end and antinode-at-open-end condition forces $L = (2n-1)\lambda/4$, which gives only odd-numbered harmonics. Even harmonics require antinode at both ends (open-open condition).
Q1 (3 marks): At the closed end, air displacement is zero, a displacement node. At the open end, air is free to move, a displacement antinode. The only standing waves that satisfy node-at-closed, antinode-at-open are those where $L = \lambda/4, 3\lambda/4, 5\lambda/4, \ldots$, odd multiples of a quarter wavelength. These give $f = v/4L, 3v/4L, 5v/4L$, only odd harmonics.
Q2 (3 marks): $f_1 = 340/(2×0.66) = 257.6$ Hz; $f_2 = 515$ Hz; $f_3 = 773$ Hz.
Q3 (4 marks): (a) Halving length doubles fundamental: $f_1' = v/(4×L/2) = v/(2L) = 2f_1$. (b) The pipe is still closed-open, so boundary conditions still require node at closed end and antinode at open end. Only odd harmonics are allowed regardless of length, so the set of harmonics remains $f_1', 3f_1', 5f_1', \ldots$
The 2013 UNSW study confirmed that a ~1.5 m Yolŋu didgeridoo produces $f_1 \approx 65$ Hz and a first overtone at 195 Hz (= 3×65 Hz). The 2nd harmonic at 130 Hz is absent, because a closed-open pipe only allows odd harmonics: $L = (2n-1)\lambda/4$, so the 2nd harmonic ($n = 1.5$, not an integer) cannot form a standing wave that satisfies both boundary conditions (node at closed end, antinode at open).
Your Think First prediction about an open-open pipe was: the longest wavelength fitting is $\lambda_1 = 2L$, giving $f_1 = v/2L$. That is correct for an open-open pipe. The didgeridoo (closed-open) gives $\lambda_1 = 4L$ and $f_1 = v/4L$, half the frequency for the same length.
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