Exercises — Harmonics and overtones — on strings and in pipes

What to see. Look at figure s01. The number of humps (antinodes) equals the harmonic number $n$ for a fixed–fixed string, because each hump is a half-wavelength loop and $L=n\lambda /2$. Answer: 3 humps $\Rightarrow$ the 3rd harmonic ($n=3$). Nodes. End clamps are always nodes (fixed = no motion). Between $n$ loops there are $n-1$ interior nodes where neighbouring loops meet. $n-1=3-1=\boxed{2\text{ interior nodes}}$

Rule of the boundary. A closed end forces a node (air can't move through the wall). An open end forces an antinode (air swings freely). See the closed-pipe column of figure s02. Fundamental. The shortest fit that puts a node at the closed end and an antinode at the open end is the distance from a node to its nearest antinode, which is a quarter wavelength: $L=\frac{\lambda }{4}\Rightarrow \boxed{\frac{1}{4}\text{ of a wavelength fits}}.$

False. "Overtone" = the next allowed frequency above $f_1$. "Harmonic" = an integer multiple $n{f}_1$. They coincide only when every integer is allowed (string, open pipe). In a closed pipe the 2nd harmonic ($2f_1$) is forbidden, so the 1st overtone is the 3rd harmonic ($3f_1$).

Tool. Fixed–fixed $\Rightarrow f_n=nv/2L$. Why $2L$? the fundamental fits one node–node loop, a half wavelength, so ${\lambda }_1=2L$ and $f_1=v/{\lambda }_1=v/2L$. $f_1=\frac{480}{2(0.80)}=\frac{480}{1.60}=300\text{ Hz}$ All harmonics are allowed, so multiply: $f_2=2f_1=600\text{ Hz},f_4=4f_1=1200\text{ Hz}.$

Tool. Open–open uses the same spacing as the string (antinode-to-antinode is also $\lambda /2$), so $f_n=nv/2L$. $f_1=\frac{340}{2(0.34)}=\frac{340}{0.68}=500\text{ Hz}.$ Every harmonic is present, so the 1st overtone is the 2nd harmonic: $f_{\text{1st overtone}}=2f_1=1000\text{ Hz}.$

Tool. Closed one end $\Rightarrow f_n=(2n-1)v/4L$, odd multiples only. Why $4L$? node–antinode needs only a quarter wave, so ${\lambda }_1=4L$ — twice as long as the open pipe's fundamental wavelength, hence half the frequency. $f_1=\frac{340}{4(0.34)}=\frac{340}{1.36}=250\text{ Hz}.$ The next allowed mode uses $n=2\Rightarrow (2\cdot 2-1)=3$: $f_{\text{1st overtone}}=3f_1=750\text{ Hz}.$ Notice: the closed fundamental ($250$) is exactly half the open fundamental ($500$) — an octave lower, matching the parent note.

Let the open fundamental be $f_1^o=v/2L$. Then the closed fundamental is $f_1^c=v/4L=\frac{1}{2}{f}_1^o$.

• Open frequencies: $f_1^o,2f_1^o,3f_1^o,\cdots =1,2,3,4,\dots$ (in units of $f_1^o$).
• Closed frequencies: odd multiples of $f_1^c=\frac{1}{2}{f}_1^o$, i.e. $1\cdot \frac{1}{2},3\cdot \frac{1}{2},5\cdot \frac{1}{2},\cdots =\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\dots$ (in units of $f_1^o$). Overlap? The open list is whole numbers $\{1,2,3,\dots \}$; the closed list is half-odd-integers $\{0.5,1.5,2.5,\dots \}$. No number is in both. The two pipes share none of their frequencies. (This is exactly the "false forecast" from the parent's example 4.)

What the finger does. Forcing a node at the centre means the string must vibrate in shapes that already have a node there — that is, in even-loop patterns. The lowest such pattern is two loops, i.e. the 2nd harmonic of the original string. $f_{\text{new lowest}}=2f_1=400\text{ Hz}.$ Picture check (s01, two-hump shape): the middle of a two-loop pattern is a node, so the boundary condition is satisfied — the string is happy vibrating there.

Set the fundamentals equal. $f_1^c=\frac{v}{4L_c},f_1^o=\frac{v}{2L_o},f_1^c=f_1^o.$ Cancel $v$ and cross-multiply: $\frac{1}{4L_c}=\frac{1}{2L_o}\Rightarrow 2L_o=4L_c\Rightarrow L_c=\frac{1}{2}{L}_o.$ A closed pipe only needs to be half as long to hit the same pitch — the whole point of stopped organ pipes saving space.

Chain the tools. First the string speed comes from tension: $v=\sqrt{T/\mu }$ (see Speed of a wave on a string). Second, the fundamental is $f_1=v/2L$. Combine: $f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}.$ Solve for $T$. Multiply by $2L$, square, then multiply by $\mu$: $(2L{f}_1{)}^2=\frac{T}{\mu }\Rightarrow T=\mu (2L{f}_1{)}^2.$ Plug in $2L{f}_1=2(0.65)(196)=254.8$ m/s: $T=(6.0\times 10^{-3})(254.8{)}^2=6.0\times 10^{-3}\times 64923\approx 389.5\text{ N}.$ So about $\boxed{390\text{ N}}$ of tension.

Tool. $f_1=v/4L$; only $v$ changes with the gas (see Speed of sound in gases). $f_1^{\text{air}}=\frac{340}{4(0.25)}=\frac{340}{1.0}=340\text{ Hz}.$ $f_1^{\text{He}}=\frac{965}{1.0}=965\text{ Hz}.$ Ratio. $\frac{f_1^{\text{He}}}{f_1^{\text{air}}}=\frac{965}{340}\approx \mathrm{2.84.}$ The pitch jumps by roughly $2.84\times$ — the same geometry, but faster sound raises every mode. This is the "helium voice" effect: your vocal-tract resonances stay the same length, but $v$ is larger, so all resonant frequencies rise.

Tool 1 — fundamentals. Open pipes: $f=v/2L$. $f_1=\frac{340}{2(0.50)}=340\text{ Hz},f_2=\frac{340}{2(0.51)}=\frac{340}{1.02}\approx 333.33\text{ Hz}.$ Tool 2 — beats. When two close frequencies overlap, the loudness throbs at the difference frequency (see Beats and resonance): $f_{\text{beat}}=|f_1-f_2|=340-333.33=\boxed{6.67\text{ Hz}}.$ You'd hear about 6–7 pulses per second.

(a) New fundamental. Sealing one end changes $v/2L\to v/4L$, halving the fundamental at fixed $L$ and $v$: $f_1^{\text{closed}}=\frac{1}{2}{f}_1^{\text{open}}=\frac{1}{2}(440)=220\text{ Hz}.$ (b) Closed ladder. Odd multiples of $220$: $220,660,1100,1540,1980,\dots \text{ Hz.}$ Compare with the old open overtones $\{880,1320,1760\}$:

• $880$? not in the closed list — gone.
• $1320$? not in the list — gone.
• $1760$? not in the list — gone. So none of the old overtones survive; the entire ladder is rebuilt. The genuinely new mode between $220$ and $660$: the closed list has $220$ then jumps straight to $660$ with nothing between, so the answer is that the new fundamental $220$ Hz itself is the new tone in that band, and the band $(220,660)$ is otherwise empty — a musical "gap" the open pipe never had (the open pipe had $440$ sitting right in that gap). Physical story: closing the end deepens the pitch by an octave and thins the spectrum (only odd harmonics), which is exactly why a stopped pipe has a hollow, "covered" timbre — see Timbre and Fourier synthesis.

Key idea. A bead disturbs the motion unless it sits exactly on a node (a point that never moves). So we must locate the nodes of the 4th harmonic. Node positions. The $n$th harmonic has $n$ loops; nodes occur where the pattern crosses zero. For the $n$th mode the displacement shape is proportional to $\sin (n\pi x/L)$, which is zero when $n\pi x/L=k\pi$, i.e. $x=\frac{k}{n}L,k=0,1,2,\dots ,n.$ Picture check (s01, four-hump shape): count the zero-crossings. For $n=4$: $x=0,\frac{1}{4}L,\frac{1}{2}L,\frac{3}{4}L,L.$ Excluding the two fixed ends ($x=0$ and $x=L$, which are trivially nodes), the interior safe spots are $\boxed{x=\frac{1}{4}L,\frac{1}{2}L,\frac{3}{4}L.}$

Set up. Closed pipe: allowed frequencies are $(2n-1)f_1$ with $f_1=v/4L$. The "3rd harmonic" means the mode whose multiplier is $3$, i.e. $n=2$ (since $2n-1=3$). So $f_{\text{3rd}}=3f_1=3\cdot \frac{v}{4L}=\frac{3v}{4L}.$ Solve for $L$. Set equal to $1500$ and rearrange: $1500=\frac{3(340)}{4L}\Rightarrow L=\frac{3(340)}{4(1500)}=\frac{1020}{6000}=0.17\text{ m}.$ Verify. $f_1=340/(4\cdot 0.17)=340/0.68=500$ Hz. Odd multiples: $500,1500,2500,\dots \text{ Hz.}$ The 3rd harmonic is indeed $1500$ Hz. ✓

Fixed–fixed string with 5 humps — which harmonic? ::: The 5th harmonic ($n=5$). Interior nodes for the $n$th harmonic of a fixed–fixed string? ::: $n-1$. Closed pipe, "3rd harmonic" — what multiplier of $f_1$? ::: $3f_1$ (not $5f_1$). Beat frequency of two tones $f_1,f_2$? ::: $|f_1-f_2|$. Tension for a target $f_1$ on a string? ::: $T=\mu (2L{f}_1{)}^2$. Sealing one end of an open pipe does what to $f_1$? ::: Halves it (drops an octave).