Exercises — Harmonics and overtones — on strings and in pipes

Kya dekhna hai. Figure s01 dekho. Humps (antinodes) ki sankhya harmonic number $n$ ke barabar hoti hai ek fixed–fixed string ke liye, kyunki har hump ek half-wavelength loop hai aur $L=n\lambda /2$. Jawab: 3 humps $\Rightarrow$ 3rd harmonic ($n=3$). Nodes. End clamps hamesha nodes hote hain (fixed = koi motion nahi). $n$ loops ke beech $n-1$ interior nodes hote hain jahan adjacent loops milte hain. $n-1=3-1=\boxed{2\text{ interior nodes}}$

Boundary ka rule. Closed end ek node force karta hai (air wall ke through move nahi kar sakti). Open end ek antinode force karta hai (air freely swing karti hai). Figure s02 ka closed-pipe column dekho. Fundamental. Sabse chhota fit jo closed end par node aur open end par antinode rakhta hai, woh node se nearest antinode tak ki doori hai, jo ek quarter wavelength hai: $L=\frac{\lambda }{4}\Rightarrow \boxed{\frac{1}{4}\text{ of a wavelength fits}}.$

Jhooth. "Overtone" = $f_1$ ke upar agla allowed frequency. "Harmonic" = ek integer multiple $n{f}_1$. Yeh tabhi same hote hain jab har integer allowed ho (string, open pipe). Closed pipe mein 2nd harmonic ($2f_1$) forbidden hai, isliye 1st overtone 3rd harmonic ($3f_1$) hota hai.

Tool. Fixed–fixed $\Rightarrow f_n=nv/2L$. $2L$ kyun? Fundamental mein ek node–node loop fit hota hai, jo half wavelength hai, isliye ${\lambda }_1=2L$ aur $f_1=v/{\lambda }_1=v/2L$. $f_1=\frac{480}{2(0.80)}=\frac{480}{1.60}=300\text{ Hz}$ Saare harmonics allowed hain, toh multiply karo: $f_2=2f_1=600\text{ Hz},f_4=4f_1=1200\text{ Hz}.$

Tool. Open–open mein same spacing use hoti hai jaise string mein (antinode-to-antinode bhi $\lambda /2$ hai), isliye $f_n=nv/2L$. $f_1=\frac{340}{2(0.34)}=\frac{340}{0.68}=500\text{ Hz}.$ Har harmonic present hai, isliye 1st overtone 2nd harmonic hai: $f_{\text{1st overtone}}=2f_1=1000\text{ Hz}.$

Tool. Closed one end $\Rightarrow f_n=(2n-1)v/4L$, sirf odd multiples. $4L$ kyun? node–antinode ke liye sirf ek quarter wave chahiye, isliye ${\lambda }_1=4L$ — open pipe ke fundamental wavelength se double, isliye frequency half. $f_1=\frac{340}{4(0.34)}=\frac{340}{1.36}=250\text{ Hz}.$ Agla allowed mode $n=2\Rightarrow (2\cdot 2-1)=3$ use karta hai: $f_{\text{1st overtone}}=3f_1=750\text{ Hz}.$ Notice: closed fundamental ($250$) exactly half hai open fundamental ($500$) ka — ek octave neeche, jo parent note se match karta hai.

Open fundamental ko $f_1^o=v/2L$ maano. Tab closed fundamental hai $f_1^c=v/4L=\frac{1}{2}{f}_1^o$.

• Open frequencies: $f_1^o,2f_1^o,3f_1^o,\cdots =1,2,3,4,\dots$ ($f_1^o$ ki units mein).
• Closed frequencies: $f_1^c=\frac{1}{2}{f}_1^o$ ke odd multiples, yaani $1\cdot \frac{1}{2},3\cdot \frac{1}{2},5\cdot \frac{1}{2},\cdots =\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\dots$ ($f_1^o$ ki units mein). Overlap? Open list mein whole numbers hain $\{1,2,3,\dots \}$; closed list mein half-odd-integers hain $\{0.5,1.5,2.5,\dots \}$. Koi number dono mein nahi hai. Dono pipes apni koi bhi frequencies share nahi karte. (Yeh exactly parent ke example 4 ka "false forecast" hai.)

Finger kya karta hai. Centre par node force karna matlab string ko un shapes mein vibrate karna hoga jinmein wahan pehle se node ho — yaani even-loop patterns mein. Sabse chhota aisa pattern do loops ka hai, yaani original string ka 2nd harmonic. $f_{\text{new lowest}}=2f_1=400\text{ Hz}.$ Picture check (s01, two-hump shape): do-loop pattern ka middle ek node hai, isliye boundary condition satisfy hoti hai — string wahan khushi se vibrate karti hai.

Fundamentals equal rakho. $f_1^c=\frac{v}{4L_c},f_1^o=\frac{v}{2L_o},f_1^c=f_1^o.$ $v$ cancel karo aur cross-multiply karo: $\frac{1}{4L_c}=\frac{1}{2L_o}\Rightarrow 2L_o=4L_c\Rightarrow L_c=\frac{1}{2}{L}_o.$ Ek closed pipe ko sirf aadha lambi honi chahiye same pitch paane ke liye — yahi stopped organ pipes ke space bachane ka poora point hai.

Tools chain karo. Pehle string speed tension se aati hai: $v=\sqrt{T/\mu }$ (dekho Speed of a wave on a string). Doosra, fundamental hai $f_1=v/2L$. Combine karo: $f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}.$ $T$ ke liye solve karo. $2L$ se multiply karo, square karo, phir $\mu$ se multiply karo: $(2L{f}_1{)}^2=\frac{T}{\mu }\Rightarrow T=\mu (2L{f}_1{)}^2.$ $2L{f}_1=2(0.65)(196)=254.8$ m/s plug karo: $T=(6.0\times 10^{-3})(254.8{)}^2=6.0\times 10^{-3}\times 64923\approx 389.5\text{ N}.$ Toh lagbhag $\boxed{390\text{ N}}$ ka tension.

Tool. $f_1=v/4L$; sirf $v$ gas ke saath badalta hai (dekho 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. $\frac{f_1^{\text{He}}}{f_1^{\text{air}}}=\frac{965}{340}\approx \mathrm{2.84.}$ Pitch roughly $2.84\times$ jump karti hai — same geometry, lekin faster sound har mode ko raise karta hai. Yeh "helium voice" effect hai: tumhari vocal-tract resonances same length par rehti hain, lekin $v$ bada hai, isliye saari resonant frequencies badhti hain.

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. Jab do close frequencies overlap karti hain, loudness difference frequency par throb karti hai (dekho Beats and resonance): $f_{\text{beat}}=|f_1-f_2|=340-333.33=\boxed{6.67\text{ Hz}}.$ Tum lagbhag 6–7 pulses per second sunoge.

(a) Naya fundamental. Ek end seal karna $v/2L\to v/4L$ change karta hai, fixed $L$ aur $v$ par fundamental ko half karta hai: $f_1^{\text{closed}}=\frac{1}{2}{f}_1^{\text{open}}=\frac{1}{2}(440)=220\text{ Hz}.$ (b) Closed ladder. $220$ ke odd multiples: $220,660,1100,1540,1980,\dots \text{ Hz.}$ Purane open overtones $\{880,1320,1760\}$ se compare karo:

• $880$? closed list mein nahi — gone.
• $1320$? list mein nahi — gone.
• $1760$? list mein nahi — gone. Toh purane overtones mein se koi bhi survive nahi karta; puri ladder rebuild hoti hai. $220$ aur $660$ ke beech genuinely naya mode: closed list mein $220$ hai phir seedha $660$ aata hai beech mein kuch nahi, isliye jawab yeh hai ki naya fundamental $220$ Hz khud us band mein naya tone hai, aur band $(220,660)$ baaki empty hai — ek musical "gap" jo open pipe mein kabhi nahi tha (open pipe mein $440$ us gap ke beech mein tha). Physical story: end close karna pitch ko ek octave deepens karta hai aur spectrum thin karta hai (sirf odd harmonics), jo exactly explain karta hai ki stopped pipe ka timbre hollow aur "covered" kyun hota hai — dekho Timbre and Fourier synthesis.

Key idea. Ek bead motion ko disturb karta hai jab tak woh exactly ek node par na ho (ek aisi jagah jo kabhi move nahi hoti). Isliye hume 4th harmonic ke nodes locate karne hain. Node positions. $n$th harmonic mein $n$ loops hote hain; nodes wahan occur karte hain jahan pattern zero cross karta hai. $n$th mode ke liye displacement shape $\sin (n\pi x/L)$ ke proportional hai, jo zero hai jab $n\pi x/L=k\pi$, yaani $x=\frac{k}{n}L,k=0,1,2,\dots ,n.$ Picture check (s01, four-hump shape): zero-crossings count karo. $n=4$ ke liye: $x=0,\frac{1}{4}L,\frac{1}{2}L,\frac{3}{4}L,L.$ Dono fixed ends ($x=0$ aur $x=L$, jo trivially nodes hain) exclude karke, interior safe spots hain: $\boxed{x=\frac{1}{4}L,\frac{1}{2}L,\frac{3}{4}L.}$

Setup. Closed pipe: allowed frequencies $(2n-1)f_1$ hain jahan $f_1=v/4L$. "3rd harmonic" matlab woh mode jiska multiplier $3$ hai, yaani $n=2$ (kyunki $2n-1=3$). Toh $f_{\text{3rd}}=3f_1=3\cdot \frac{v}{4L}=\frac{3v}{4L}.$ $L$ ke liye solve karo. $1500$ ke equal karo aur rearrange karo: $1500=\frac{3(340)}{4L}\Rightarrow L=\frac{3(340)}{4(1500)}=\frac{1020}{6000}=0.17\text{ m}.$ Verify karo. $f_1=340/(4\cdot 0.17)=340/0.68=500$ Hz. Odd multiples: $500,1500,2500,\dots \text{ Hz.}$ 3rd harmonic sach mein $1500$ Hz hai. ✓

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).