Exercises — Beats — derivation, applications

(a) HUM KYA KARTE HAIN: do frequencies ko subtract karo aur sign drop kar do. $f_{\text{beat}}=|250-254|=4\text{ Hz}$ KYUN: ek beat ek complete loud→soft→loud cycle hota hai, aur wo cycles exactly us rate par hoti hain jis par do waves drift apart aur re-line-up karti hain — jo frequency ka difference hai. (b) Pitch average hai: $f_{\text{avg}}=\frac{250+254}{2}=252\text{ Hz}$ Answer: 4 beats/s, 252 Hz note throbbing hai.

(a) FALSE. Same frequency $\Rightarrow f_{\text{beat}}=|f-f|=0$. Zero beats ka matlab hai loudness kabhi swell aur fade nahi hoti — ek perfectly steady note. (Yahi "tuning finish ho gayi" ki condition hai.) (b) FALSE. $60$ Hz par swelling ek second mein 60 baar hoti hai — ear ke liye follow karna bahut fast hai; tum bas do rough alag tones sunte ho. $3$ Hz par tum ek slow, clearly countable waah…waah…waah sunte ho. Beats sirf tab audible hoti hain jab $|f_1-f_2|$ small ho (roughly $10$ Hz se kam).

(a) $f_{\text{beat}}=|440-436|=4$ Hz. (b) WHAT/WHY: ek silence har beat cycle mein ek baar hoti hai, isliye silences ke beech ka gap beat period $T_{\text{beat}}=1/f_{\text{beat}}$ hai — reciprocal, kyunki "4 events per second" ka matlab hai "har $1/4$ second mein ek event". $T_{\text{beat}}=\frac{1}{4}=0.25\text{ s}$ Answer: 4 Hz; har $0.25$ s mein ek silence.

Step 1 — beats per second: $f_{\text{beat}}=\frac{10\text{ beats}}{2\text{ s}}=5$ Hz. Step 2 — absolute value ko undo karo. Formula $f_{\text{beat}}=|f_{\text{fork}}-500|$ ek sign hide karta hai: $|x|=5$ ke do solutions hain, $x=+5$ ya $x=-5$. $f_{\text{fork}}=500+5=505\text{ Hz}\text{ya}{f}_{\text{fork}}=500-5=495\text{ Hz}$ Answer: $495$ Hz ya $505$ Hz. (Hum abhi nahi bata sakte kaunsa hai — yahi ambiguity L3 ka pura point hai.)

Step 1 — do candidates. $|f_X-320|=4\Rightarrow f_X=316$ ya $324$ Hz. Step 2 — wax kya karta hai. Wax mass $m$ add karta hai. Fork ke liye jo Simple Harmonic Motion ki tarah behave karta hai, $f=\frac{1}{2\pi }\sqrt{k/m}$, isliye zyaada mass ⇒ kam $f$. Waxing ke baad, $f_X$ girta hai. Step 3 — har candidate ko observation "beats 4 → 6 gayi" (320 se aur dur) ke against test karo:

• Agar $f_X=324$: waxing ise 320 ki taraf neeche le jaati hai → gap shrinks → beats decrease hongi. ✗ observation ke saath contradict karta hai.
• Agar $f_X=316$: waxing ise 320 se door neeche le jaati hai → gap grows → beats increase hoti hain. ✓ "6 tak badh gayi" se match karta hai. Answer: $f_X=316$ Hz. (Sanity check: waxing ke baad yeh $320-6=314$ Hz par hai, aur $314<316$ — haan, yeh neeche gaya, added mass ke saath consistent hai.)

Step 1: $|f_Y-384|=3\Rightarrow f_Y=381$ ya $387$ Hz. Step 2 — filing kya karta hai. Filing mass remove karti hai, isliye $m$ decreases aur $f=\frac{1}{2\pi }\sqrt{k/m}$ increase karta hai. Filing ke baad, $f_Y$ upar jaata hai. Step 3 — har candidate ko "beats 3 → 1 gayi" (384 ke karib) ke against test karo:

• Agar $f_Y=381$: filing ise 384 ki taraf upar uthati hai → gap shrinks → beats decrease hoti hain. ✓
• Agar $f_Y=387$: filing ise 384 se door upar uthati hai → gap grows → beats increase hoti hain. ✗ Answer: $f_Y=381$ Hz. Moral yeh hai: wax lowers, filing raises; beat count jis bhi direction mein move kare woh batata hai ki unknown standard ke kis side se start kiya. Woh single observation $\pm$ sign resolve kar deta hai.

(a) "Sent" aur "received" ko do superposing signals ki tarah treat karo: $f_{\text{beat}}=|2000-2012|=12\text{ Hz}$ (b) Doppler Effect ke according, ek approaching reflector wave crests ko compress karta hai jo woh bounce back karta hai, isliye har returning crest jaldi aata hai → higher frequency. (Ek receding target ek lower returned frequency deta aur equally sized beat tabhi deta agar shift magnitude equal hoti.) $12$ Hz beat machine ka direct, countable read-out hai submarine ki speed ka.

(a) $y_1=a\cos (2\pi \cdot 500t)$, $y_2=a\cos (2\pi \cdot 508t)$ ke saath, $\cos C+\cos D=2\cos \frac{C-D}{2}\cos \frac{C+D}{2}$ apply karo (yeh wahi Superposition Principle algebra hai jo Standing Waves ke liye use hoti hai, bas space ki jagah time ke saath): $y(t)=2a\underset{\text{slow envelope}}{\underbrace{\cos (2\pi \cdot 4t)}}\underset{\text{fast heard tone}}{\underbrace{\cos (2\pi \cdot 504t)}}$ Yahan $\frac{508-500}{2}=4$ aur $\frac{508+500}{2}=504$. (b) Envelope frequency $=4$ Hz. Lekin loudness depend karta hai $|A(t)|=|2a\cos (2\pi \cdot 4t)|$ par, aur $|\cos |$ ek envelope cycle mein do baar peak hit karta hai (ek baar zero ke upar hump ke liye, ek baar neeche dip ke liye). Isliye: $f_{\text{beat}}=2\times 4=8\text{ Hz}=|500-508|.✓$

Step 1 — A ko anchor ki tarah use karo. A sabse zyaada hai, isliye B aur C dono A se neeche hain.

• A–B beat $=5$ aur B, A se neeche hai $\Rightarrow f_B=260-5=255$ Hz.
• A–C beat $=8$ aur C, A se neeche hai $\Rightarrow f_C=260-8=252$ Hz. Step 2 — B–C beat ke saath consistency check karo. $|f_B-f_C|=|255-252|=3\text{ Hz.}✓$ Teeno measurements mutually consistent hain, picture confirm karti hai. Answer: $f_B=255$ Hz, $f_C=252$ Hz. "Agar $2$ Hz hota" wala twist: Agar A–C $2$ Hz hota to $f_C=260-2=258$ Hz, giving $|f_B-f_C|=|255-258|=3$ Hz — phir bhi $3$-beat measurement ke saath consistent! Isliye B–C beat akela in dono worlds ko distinguish nahi kar sakta; extra information "A sabse zyaada hai" plus A–C beat hi frequencies pin karta hai. Isliye real tuning mein ek known reference use hota hai: anchor ke bina beat differences ek web banaate hain jiske multiple sign-solutions hote hain.

Step 1 — Saath mein bajao, beats count karo. Maan lo tum $n$ beats/s sunte ho. To $|f_{\text{unknown}}-256|=n$, isliye unknown $256+n$ ya $256-n$ hai. Kyun: absolute value do candidates deta hai; yeh gap ka size measure karta hai. Step 2 — Unknown fork par wax lagao, phir bajao, note karo ki beats badhti hain ya ghatti hain. Wax mass $m$ add karta hai, aur $f=\frac{1}{2\pi }\sqrt{k/m}$ zyaada mass ke saath girta hai. Isliye waxing ke baad $f_{\text{unknown}}$ decrease hota hai. Step 3 — Side decide karo:

• Agar beats decrease hoti hain, fork 256 ki taraf move hua, isliye yeh 256 se upar tha → $f_{\text{unknown}}=256+n$.
• Agar beats increase hoti hain, fork 256 se door gaya, isliye yeh 256 se neeche tha → $f_{\text{unknown}}=256-n$. Step 4 — Worked instance. Maan lo Step 1 mein $n=6$ beats/s milti hain aur Step 2 mein beats increase hokar $\approx 8$/s ho jaati hain. To fork 256 se neeche hai: $f_{\text{unknown}}=256-6=250\text{ Hz}.$ Yeh kyun kaam karta hai: wax invisible $\pm$ sign ko ek visible countable rhythm ki change mein convert karta hai. Yeh wahi comparator idea hai jo heterodyne radio aur Amplitude Modulation ke peeche hai — ek tiny frequency difference ko ek slow, measurable beat ki tarah read karo.

Beat frequency difference $|f_1-f_2|$ hai; jo pitch tum sunte ho woh average hai; wax/file test wo sign resolve karta hai jo absolute value hide karta hai.