Sound intensity — decibels (logarithmic scale): Worked examples

1.6.23 · D3 · Physics › Oscillations & Waves › Sound intensity — decibels (logarithmic scale)

Definition Woh ek formula jo is page par sab kuch use karta hai (recap)

Sound intensity $I$ power per unit area hai, $I = P / A$ , measured in $W/m^{2}$ . Decibel level $I$ ko ek fixed reference se compare karta hai: $β = 10 lo g_{10} (\frac{I}{I _{0}}) dB, I_{0} = 1 0^{- 12} W/m^{2}$

$I_{0}$ kahan se aata hai: yeh threshold of hearing hai — woh sabse quiet sound jo ek human 1 kHz par detect kar sakta hai. Hum har sound ko is floor ke relative measure karte hain, isliye sabse quiet audible sound $lo g_{10} (1) = 0 dB$ padhta hai.
Log kyun: audible range ek factor of $\sim 1 0^{12}$ span karta hai; ek logarithm factors of ten count karta hai aur use $0$ – $120$ mein compress kar deta hai. (Dekho Logarithms and exponentials.)
"Bel" kya hai: $lo g_{10} (I / I_{0})$ akele level in bels mein hai — ek bel = intensity mein ek poora factor of ten. Bel ek coarse step hai, isliye hum decibels = tenths of a bel mein kaam karte hain, aur yahi wajah hai $\times 10$ front par aata hai. Isliye "deci-bel" (dB).
Difference rule (neeche baar baar use hoga): $Δ β = β_{2} - β_{1} = 10 lo g_{10} (I_{2} / I_{1})$ ; $I_{0}$ cancel ho jaata hai, isliye tumhe sirf ratio chahiye.

The scenario matrix

Har decibel problem in case classes mein se ek (ya combination) hai. Right-hand column us example ka naam deta hai jo use cover karta hai.

Cell	Case class	Tricky kyun hai	Covered by
A	Intensity → dB (forward)	bas plug in karo, exponents dekho	Ex 1
B	dB → intensity (log ko invert karo)	×10 undo karo, phir $lo g$ undo karo	Ex 2
C	Sirf ratio — sources add hote hain	$I_{0}$ cancels; intensities add hoti hain, dB nahi	Ex 3
D	Distance / inverse-square	ratio $(r_{1} / r_{2})^{2}$ se aata hai	Ex 4
E	Degenerate: $I = I_{0}$ aur $I < I_{0}$	zero dB, phir negative dB	Ex 5
F	Pressure diya hai, intensity nahi	$20 lo g_{10}$ use karo kyunki $I \propto p^{2}$	Ex 6
G	Word problem: distance aur counting combine karo	do ratios chain karo	Ex 7
H	Exam twist: "kitne quiet sources = ek loud one?"	log ke andar count ke liye solve karo	Ex 8
I	Limiting behaviour: $I \to 0$	$β \to - \infty$ ; silence ka "matlab"	Ex 9

Teen signs notice karo jo $β$ le sakta hai: positive (threshold se zyada loud, usual case), zero (exactly threshold), aur negative (reference se quieter — real, bas $I_{0}$ ke neeche). Ex 5 aur Ex 9 ensure karte hain ki tumne teeno dekhe hain.

Example 1 — Cell A · Intensity → dB

Worked example Ek quiet library mein

$I = 3.2 \times 1 0^{- 9} W/m^{2}$ hai. $β$ nikalo. Forecast: aage padhne se pehle dB guess karo. Threshold $1 0^{- 12}$ hai, yeh kuch hazaar times bada hai, toh expect karo kuch 30s mein.

Step 1. Ratio banao $\frac{I}{I _{0}} = \frac{3.2 \times 1 0 ^{- 9}}{1 0 ^{- 12}} = 3.2 \times 1 0^{3} = 3200$ . Yeh step kyun? $β$ sirf isse matlab rakhta hai ki threshold se kitne times upar ho — ek pure number. $1 0^{- 9}$ ko $1 0^{- 12}$ se divide karne ka matlab exponents subtract karna: $- 9 - (- 12) = + 3$ .

Step 2. Log lo: $lo g_{10} (3200) = lo g_{10} (3.2) + 3 = 0.505 + 3 = 3.505$ . Yeh step kyun? Log factors of ten count karta hai. Humne $3200 = 3.2 \times 1 0^{3}$ split kiya taaki $1 0^{3}$ exactly $3$ contribute kare aur sirf leading $3.2$ ko lookup chahiye.

Step 3. 10 se multiply karo: $β = 10 \times 3.505 = 35.05 \approx 35 dB$ . Yeh step kyun? Decibels tenths of a bel hain, isliye formula ×10 carry karta hai.

Verify: $35 dB$ whisper (20) aur conversation (60) ke beech hai — library ke liye bilkul sahi, aur hamare "mid-30s" forecast se match karta hai. ✓

Example 2 — Cell B · dB → intensity

Worked example Ek rock concert

$110 dB$ ka hai. Actual intensity $I$ kya hai? Forecast: $110$ , $120 dB$ pain point ( $I = 1$ ) se $10$ neeche hai. Ten dB down matlab ÷10, toh guess karo $I \approx 0.1 W/m^{2}$ .

Step 1. ×10 undo karo: $lo g_{10} \frac{I}{I _{0}} = \frac{110}{10} = 11$ . Yeh step kyun? Formula ne last mein 10 se multiply kiya tha; backwards chalte hain toh pehle 10 se divide karo.

Step 2. Log undo karo: $\frac{I}{I _{0}} = 1 0^{11}$ . Yeh step kyun? $lo g_{10} (x)$ ka inverse $1 0^{x}$ hai — dekho Logarithms and exponentials. "Kaunse number ka log $11$ hai?" → $1 0^{11}$ .

Step 3. $I_{0}$ se wapas multiply karo: $I = 1 0^{11} \times 1 0^{- 12} = 1 0^{- 1} = 0.1 W/m^{2}$ . Yeh step kyun? Humne shuru mein $I_{0}$ se divide kiya tha; real intensity with units recover karne ke liye wapas multiply karo.

Verify: Forward plug karo — $10 lo g_{10} (0.1/1 0^{- 12}) = 10 lo g_{10} (1 0^{11}) = 110 dB$ . ✓ Forecast se match karta hai.

Example 3 — Cell C · Sources add hote hain (sirf ratio)

Worked example Ek akela cricket

$42 dB$ par chirp karta hai. Ek field mein paanch identical crickets hain. Combined level? Forecast: $42 \times 5$ nahi. Paanch times intensity ek factor of ten se kam hai, toh expect karo roughly $+ 7 dB$ → roughly $49 dB$ .

Step 1. Paanch identical sources → intensities add → $I_{new} = 5 I_{one}$ , toh ratio $5$ hai. Yeh step kyun? Energy flows add hoti hain (dekho Wave energy and power); dB add nahi hote, kyunki woh logarithms hain.

Step 2. Difference rule use karo — $I_{0}$ cancel ho jaata hai: $Δ β = 10 lo g_{10} (5) = 10 \times 0.699 = 6.99 dB$ . Yeh step kyun? Humein sirf ratio $5$ pata hai, har intensity nahi. Difference rule exactly iske liye bani hai — "ratio in, dB-change out".

Step 3. Original mein add karo: $β = 42 + 6.99 \approx 49.0 dB$ .

Verify: Sanity — doubling $+ 3$ deta hai, ten-fold $+ 10$ deta hai; paanch ( $2$ aur $10$ ke beech) $+ 7$ deta hai, jo beech mein hai. ✓ Aur parent note ke mistake box mein bataye gaye "galat" jawaab $210 dB$ se kaafi door hai.

Example 4 — Cell D · Distance ke saath Inverse-square

Figure — Sound intensity — decibels (logarithmic scale)

Worked example Ek siren

$r_{1} = 10 m$ par $95 dB$ padhta hai. $r_{2} = 30 m$ par kya padhega (open field, no echoes)? Forecast: Distance ×3 → intensity ÷9. Yeh roughly ek factor of ten hai, toh roughly $- 9$ ya $- 10 dB$ → mid-80s.

Step 1. Geometry se intensity ratio: $\frac{I _{2}}{I _{1}} = (\frac{r _{1}}{r _{2}})^{2} = (\frac{10}{30})^{2} = \frac{1}{9}$ . Yeh step kyun? Source ki sari power $P$ usse surrounding sphere cross karti hai. Radius $r$ ka ek sphere area $A = 4 π r^{2}$ rakhta hai, isliye intensity $I = P / A = P / (4 π r^{2})$ hai — area $r^{2}$ ke saath badhta hai, isliye $I$ $1/ r^{2}$ ke saath girta hai (dekho Inverse-square law for radiation). $r$ triple karo, area ×9, intensity ÷9. Figure mein pink arc dekho: teen times radius, nine times surface.

Step 2. $Δ β = 10 lo g_{10} (\frac{1}{9}) = - 10 lo g_{10} (9) = - 10 \times 0.954 = - 9.54 dB$ . Yeh step kyun? $1$ se neeche ratio → log negative hai → level gir jaata hai, exactly jaise physics demand karti hai.

Step 3. $β_{2} = 95 - 9.54 \approx 85.5 dB$ .

Verify: Sign check — door hona matlab quieter, aur humein decrease mila. Magnitude check — ÷9 roughly ÷10 ( $- 10$ ) hai, toh $- 9.5$ sahi hai. Forecast se match karta hai. ✓

Example 5 — Cell E · Degenerate inputs ( $I = I_{0}$ aur $I < I_{0}$ )

Worked example (i) Ek sound exactly

$I = I_{0} = 1 0^{- 12} W/m^{2}$ hai. (ii) Ek sound $I = 4 \times 1 0^{- 13} W/m^{2}$ hai. Dono levels nikalo. Forecast: (i) exactly $0 dB$ hona chahiye. (ii) threshold se neeche hai, toh negative number aane wala hai.

Step 1 (i). $β = 10 lo g_{10} (\frac{I _{0}}{I _{0}}) = 10 lo g_{10} (1) = 10 \times 0 = 0 dB$ . Yeh step kyun? $lo g_{10} (1) = 0$ kyunki $1 0^{0} = 1$ : threshold se upar zero factors of ten. Yahi wajah hai scale $I_{0}$ par anchor ki gayi — sabse quiet audible sound $0$ padhta hai.

Step 2 (ii). Ratio $= \frac{4 \times 1 0 ^{- 13}}{1 0 ^{- 12}} = 0.4$ , jo 1 se kam hai. Yeh step kyun? Reference se neeche ratio $1$ se neeche jaata hai, aur $1$ se neeche number ka log negative hota hai.

Step 3 (ii). $β = 10 lo g_{10} (0.4) = 10 \times (- 0.398) = - 3.98 \approx - 4.0 dB$ . Yeh step kyun? Negative dB bilkul legal hai — iska matlab sirf "reference se quieter" hai. Tumhara kaan pakad nahi payega, lekin maths $I > 0$ ke liye defined hai.

Verify: (i) exactly $0$ ✓. (ii) $- 4 dB$ negative hai jaise forecast tha; aur $0.4 \approx \frac{1}{2.5}$ , roughly half, aur half $- 3 dB$ hai, toh $- 4$ (thoda half se kam) consistent hai. ✓

Example 6 — Cell F · Pressure diya gaya (use $20 lo g_{10}$ )

Worked example Ek microphone sound

pressure doubling measure karta hai, $p_{2} = 2 p_{1}$ . Level kitne dB badhta hai? Forecast: Pressure double karna $+ 3 dB$ nahi hai (woh intensity ke liye tha). Pressure intensity mein do baar effect daalta hai, toh expect karo $+ 6 dB$ .

Step 1. Link yaad karo $I \propto p^{2}$ from Sound waves — pressure & displacement. Toh intensity ratio hai $\frac{I _{2}}{I _{1}} = (\frac{p _{2}}{p _{1}})^{2} = 2^{2} = 4$ . Yeh step kyun? Intensity pressure ke square ke proportional hai, isliye $p$ double karna $I$ four times kar deta hai. Decibel definition intensity ke baare mein hai, isliye pehle convert karna zaroori hai.

Step 2. $Δ β = 10 lo g_{10} (4) = 10 \times 0.602 = 6.02 dB$ . Yeh step kyun? Jab intensity ratio ( $4$ ) mil gayi, ordinary $10 lo g_{10}$ rule apply hoti hai.

Shortcut (same answer). $Δ β = 10 lo g_{10} (p_{2} / p_{1})^{2} = 20 lo g_{10} (p_{2} / p_{1}) = 20 lo g_{10} (2) = 6.02 dB$ . Yeh step kyun? Log ke front se exponent $2$ nikalna $10$ ko $20$ bana deta hai — exactly yahi se " $20 lo g$ for pressure" rule aata hai.

Verify: Dono routes $6.02 dB$ dete hain. ✓ Aur yeh parent note ki teesri mistake hai, ab cleanly solve ki gayi.

Example 7 — Cell G · Distance aur counting combine karo

Worked example Ek single machine

$2 m$ par $88 dB$ padhta hai. Tum chaar identical machines install karte ho aur group se $6 m$ par khade ho. Level estimate karo (unhe ek point source treat karo, free field). Forecast: Zyada machines upar push karte hain; zyada distance neeche pull karta hai. Expect karo ki dono effects partly cancel ho jaayenge — original ke aas paas kahin.

Step 1. Counting factor: $4$ machines → intensity ×4. Yeh step kyun? Intensities add hoti hain, isliye chaar equal sources chaar times power flow dete hain.

Step 2. Distance factor: $r$ $2 \to 6$ jaata hai, woh ×3 hai, toh intensity $\times (\frac{2}{6})^{2} = \times \frac{1}{9}$ . Yeh step kyun? Inverse-square phir — ratio $(r_{1} / r_{2})^{2}$ hai.

Step 3. Do ratios chain karo (woh multiply hote hain): $\frac{I _{2}}{I _{1}} = 4 \times \frac{1}{9} = \frac{4}{9}$ . Yeh step kyun? Intensity par independent effects multiply hote hain; log ke andar, multiplication do dB changes ka sum ban jaata hai.

Step 4. $Δ β = 10 lo g_{10} (\frac{4}{9}) = 10 (lo g_{10} 4 - lo g_{10} 9) = 10 (0.602 - 0.954) = - 3.52 dB$ . Phir $β_{2} = 88 - 3.52 \approx 84.5 dB$ .

Verify: Split-check — counting akela $+ 10 lo g_{10} 4 = + 6.0$ tha, distance akela $- 10 lo g_{10} 9 = - 9.5$ tha; sum $= - 3.5$ . ✓ Forecast ke mutabiq original ke near aaya.

Example 8 — Cell H · Exam twist: count ke liye solve karo

Worked example Ek concert ko

$118 dB$ reach karna hai. Har speaker $100 dB$ deliver karta hai (akele, audience par). Kitne identical speakers $N$ chahiye? Forecast: Humein $+ 18 dB$ chahiye. Yeh $+ 10$ (×10) aur $+ 20$ (×100) ke beech hai, toh guess karo $10$ aur $100$ speakers ke beech.

Step 1. Needed gain: $Δ β = 118 - 100 = 18 dB$ . Yeh step kyun? $N$ speakers ek ki intensity ka $N \times$ dete hain; woh ratio dB raise karta hai.

Step 2. Difference rule equal set karo: $18 = 10 lo g_{10} (N)$ , toh $lo g_{10} (N) = 1.8$ . Yeh step kyun? Intensity ratio exactly count $N$ hai, isliye woh log ke andar baithta hai.

Step 3. Invert karo: $N = 1 0^{1.8} = 63.1$ . Yeh step kyun? Log undo karo 10 ko power tak raise karke. Tum $63.1$ speakers nahi khareed sakte, isliye upar $N = 64$ tak round karo taaki pakka $118$ clear ho jaye.

Verify: $N = 63.1$ exactly $118 dB$ deta hai; $64$ ke saath, $β = 100 + 10 lo g_{10} 64 = 100 + 18.06 = 118.06 dB \geq 118$ . ✓ Aur $63.1$ forecast ke mutabiq $10$ aur $100$ ke beech hai.

Example 9 — Cell I · Limiting behaviour ( $I \to 0$ )

Worked example Kya hota hai

$β$ ko jab intensity perfect silence ki taraf fade ho jaati hai, $I \to 0$ ? Forecast: Silence "infinitely below" reference hona chahiye. Expect karo $β \to - \infty$ , koi finite floor nahi.

Step 1. Ratio dekho: jaise $I \to 0^{+}$ , $\frac{I}{I _{0}} \to 0^{+}$ (ek positive number zero ki taraf shrink ho raha hai). Yeh step kyun? $I_{0}$ fixed hai; sirf top shrink karta hai.

Step 2. Log track karo: $lo g_{10} (x) \to - \infty$ jaise $x \to 0^{+}$ — Logarithms and exponentials mein graph dekho. Toh $β = 10 lo g_{10} (I / I_{0}) \to - \infty$ . Yeh step kyun? Har agla intensity ka ÷10 another $10 dB$ subtract karta hai, hamesha ke liye — koi smallest value nahi hai.

Step 3. Interpretation: true silence ke liye koi finite dB nahi hai; $β = 0$ sirf hearing threshold hai, scale ka bottom nahi.

Verify: Numerically, $I = 1 0^{- 13}$ $- 10 dB$ deta hai; $1 0^{- 15}$ $- 30$ deta hai; $1 0^{- 20}$ $- 80$ deta hai — bina kisi floor ke neeche march karta hai. ✓ $- \infty$ limit aur forecast confirm hota hai.

Recall Which cell is this? (self-test)

"Ek speaker $4 m$ par 90 dB padhta hai; $12 m$ par level nikalo." — kaun sa case class? ::: Cell D (inverse-square distance), Ex 4 ki tarah. "Pressure teen times ho jaata hai; dB rise nikalo." — kaun sa case class? ::: Cell F ( $20 lo g_{10}$ use karo), Ex 6 ki tarah. "80 dB reach karne ke liye kitne 70 dB violins chahiye?" — kaun sa case class? ::: Cell H (count ke liye solve karo), Ex 8 ki tarah. Kya $β$ negative ho sakta hai, aur iska matlab kya hai? ::: Haan — iska matlab $I < I_{0}$ hai, sound hearing reference se neeche hai (Ex 5). Ex 8 mein speaker count upar kyun round karte hain? ::: $63.1$ speakers impossible hai aur target se kam padega; $64$ guarantee karta hai ki target clear ho. "Bel" kya hai, aur hum decibels kyun use karte hain? ::: Ek bel = $lo g_{10} (I / I_{0})$ , intensity mein ek poora factor of ten — bahut coarse; ek decibel ek bel ka tenth hai, isliye ×10.