Sound intensity — decibels (logarithmic scale): Exercises

1.6.23 · D4 · Physics › Oscillations & Waves › Sound intensity — decibels (logarithmic scale)

Intuition Is page ko kaise use karein

Har problem ko pehle khud solve karo, phir solution kholo. Problems ek seedhi chadti ladder ki tarah hain: pehle sirf formula pehchanana, phir usse apply karna, phir multi-step situations ka analysis, phir kai ideas ko ek saath synthesize karna, aur aakhir mein mastery problems jo sab kuch mix karti hain. Har level ke baad ek [!mistake] callout hai jo exactly woh trap batata hai jo uss rung par zyaadatar logon ko pakadti hai.

Jo kuch bhi chahiye woh parent note mein hai: Sound intensity — decibels (logarithmic scale). Ye teen cheezein hamesha haath mein rakho — level formula, difference rule, aur reference intensity: $β = 10 lo g_{10} (\frac{I}{I _{0}}) dB, Δ β = 10 lo g_{10} (\frac{I _{2}}{I _{1}}), I_{0} = 1 0^{- 12} W/m^{2} .$

Useful log values (pehle do yaad kar lo): $lo g_{10} 2 \approx 0.301$ , $lo g_{10} 3 \approx 0.477$ , $lo g_{10} 5 \approx 0.699$ , $lo g_{10} 16 \approx 1.204$ .

Intuition Edge case —

negative dB ka matlab kya hota hai? Formula tab bhi kaam karna band nahi karta jab koi sound reference $I_{0}$ se zyaada dheemi ho. Agar $I < I_{0}$ hai, toh ratio $I / I_{0}$ ek aisi number hai jo 1 se kam hai, aur 1 se neeche ki kisi number ka logarithm negative hota hai — isliye $β$ negative aata hai.

$I = I_{0} \Rightarrow β = 10 lo g_{10} (1) = 0 dB$ (threshold, prakriti ka "zero").
$I = \frac{1}{10} I_{0} = 1 0^{- 13} W/m^{2} \Rightarrow β = 10 lo g_{10} (1 0^{- 1}) = - 10 dB$ .

Toh $0$ dB khamoshi nahi hai aur scale ka floor bhi nahi hai — yeh sirf ek marker hai jo kehta hai "ek aam insaan jo sabse dheemi awaaz sun sakta hai, utna loud." $- 10$ dB ki awaaz ek real, physical sound hai; yeh sirf itni dheemi hai ki zyaadatar kaanon ke liye sunna mushkil hai (ek bahut sensitive microphone ya ek soundproof room mein baitha keen listener phir bhi isse register kar sakta hai). Negative dB = reference se neeche, matlab "koi sound nahi" nahi. Sachchi khamoshi ( $I = 0$ ) mein $β = 10 lo g_{10} (0) = - \infty$ hoga, scale ke bilkul neeche se bhi bahar.

Prerequisites jo tum khulle rakhna chahoge: Logarithms and exponentials, Wave energy and power, Inverse-square law for radiation, Sound waves — pressure & displacement, Loudness vs intensity — psychoacoustics, Doppler effect.

Level 1 — Recognition

Goal: formula mein plug karo aur result padhke wapas lo. Abhi koi trap nahi — bas fluent bano.

Recall Solution 1.1

KYA karte hain: $I$ aur $I_{0}$ ko $β = 10 lo g_{10} (I / I_{0})$ mein daalo. KYUN: $β$ poochh raha hai "threshold se kitne factors of ten upar?", aur log unhe count karta hai. $\frac{I}{I _{0}} = \frac{1 0 ^{- 9}}{1 0 ^{- 12}} = 1 0^{- 9 - (- 12)} = 1 0^{3} .$ Exponents subtract karna ( $- 9 - (- 12) = 3$ ) batata hai ki library threshold se $3$ factors of ten upar hai. $β = 10 lo g_{10} (1 0^{3}) = 10 \times 3 = 30 dB .$

Recall Solution 1.2

KYA karte hain: formula ko step by step invert karo. KYUN: $β$ banane ke liye jo operations hue the, unhe ab ulte order mein undo karo. $70 = 10 lo g_{10} \frac{I}{I _{0}} .$ 10 se divide karo (undo the "deci" ×10): $lo g_{10} \frac{I}{I _{0}} = 7.$ Dono sides pe 10 ki power uthao (log undo karo — yeh pooch raha hai "10 ki kaunsi power yeh ratio deti hai?"): $\frac{I}{I _{0}} = 1 0^{7} \Rightarrow I = 1 0^{7} \times 1 0^{- 12} = 1 0^{- 5} W/m^{2} .$

Recall Solution 1.3

KYA karte hain: difference rule $Δ β = 10 lo g_{10} (I_{2} / I_{1})$ ko ratio $10$ ke saath use karo. KYUN difference rule: hum sirf change ki parwah karte hain, aur jab hum do levels subtract karte hain toh $I_{0}$ cancel ho jaata hai. $Δ β = 10 lo g_{10} (10) = 10 \times 1 = + 10 dB .$ Yeh headline fact hai: intensity mein har ×10 exactly +10 dB hai.

Level 2 — Application

Goal: formula ko ek physical idea ke saath combine karo (doubling, distance, ratios).

Recall Solution 2.1

KYA karte hain: intensities add hoti hain, toh do equal fans $2 I$ dete hain. Phir difference rule apply karo. KYUN intensities add hoti hain, levels nahi: intensity energy-flow density hai (power per area ka ek physical amount), aur power add hoti hai. Levels logarithms hain — kisi sum ka log, logs ke sum ke barabar nahi hota. $Δ β = 10 lo g_{10} (2) = 10 \times 0.301 = 3.01 \approx + 3 dB .$ $β_{new} = 45 + 3 = 48 dB .$

Recall Solution 2.2

KYA karte hain: $I \propto 1/ r^{2}$ use karo (from Inverse-square law for radiation) intensity ratio nikalne ke liye, phir dB mein convert karo. KYUN: ek point source apni power ko $4 π r^{2}$ area ke sphere par failaata hai; $r$ aadha karne se woh area $4$ se shrink hota hai, isliye intensity chaar guna ho jaati hai. $\frac{I _{2}}{I _{1}} = (\frac{r _{1}}{r _{2}})^{2} = (\frac{10}{5})^{2} = 2^{2} = 4.$ $Δ β = 10 lo g_{10} (4) = 10 \times (2 lo g_{10} 2) = 10 \times 0.602 = + 6.0 dB .$ $β_{new} = 58 + 6 = 64 dB .$

Neeche di gayi figure mein woh do arcs dikhaye gaye hain jinhe sound cross karta hai. $5 m$ par green arc dekho: same power $10 m$ ke blue arc ke area ke ek-chauthai mein pack ho jaati hai, isliye wahaan intensity chaar guna zyaada hai — aur chaar guna exactly wohi $+ 6 dB$ hai jo humne calculate kiya.

Figure — Sound intensity — decibels (logarithmic scale)

Recall Solution 2.3

KYA karte hain: ratio solve karne ke liye difference rule invert karo. KYUN: dB mein difference $10 lo g_{10}$ hai intensity ratio ka, toh hum log undo karte hain. $Δ β = 95 - 65 = 30 dB = 10 lo g_{10} \frac{I _{2}}{I _{1}} .$ $lo g_{10} \frac{I _{2}}{I _{1}} = 3 \Rightarrow \frac{I _{2}}{I _{1}} = 1 0^{3} = 1000 .$ 30 dB ka gap intensity mein ek hazaar guna hai — "lagbhag 1.5×" nahi.

Level 3 — Analysis

Goal: kai steps, ya ek non-power-of-ten ratio jisme real log values chahiye.

Recall Solution 3.1

KYA karte hain: distance ×4 hoti hai, toh inverse-square se intensity ÷ $4^{2} = 16$ ho jaati hai. KYUN square: $I \propto 1/ r^{2}$ , aur $(r_{1} / r_{2})^{2} = (2/8)^{2} = (1/4)^{2} = 1/16$ . $Δ β = 10 lo g_{10} \frac{1}{16} = - 10 lo g_{10} 16 = - 10 \times 1.204 = - 12.0 dB .$ Negative sign kehta hai level drop hua (hum door gaye). $β_{new} = 80 - 12 = 68 dB .$

Neeche di gayi figure level ko distance ke against ek smooth curve par plot karti hai. Blue curve ko green dot ( $2 m$ par $80 dB$ ) se neeche red dot ( $8 m$ par $68 dB$ ) tak follow karo: gray double-arrow $- 12 dB$ drop mark karta hai, aur notice karo ki curve source ke paas steeply girti hai lekin door jaake flat ho jaati hai — yeh $1/ r^{2}$ law hai jo bade $r$ par apni pakad khota hai.

Recall Solution 3.2

KYA karte hain: rise $80 - 60 = 20 dB$ hai; intensity ratio nikalo, jo $N$ ke barabar hai. KYUN $N$ : $N$ identical sources ek machine ki intensity ka $N$ guna dete hain. $20 = 10 lo g_{10} (N) \Rightarrow lo g_{10} (N) = 2 \Rightarrow N = 1 0^{2} = 100 .$ Sanity check the compression: ek sau machines sirf 20 dB zyaada karti hain ek se — yahi toh log scale ka poora point hai.

Recall Solution 3.3

KYA karte hain: " $10%$ pass hoti hai" matlab $I_{2} / I_{1} = 0.10 = 1 0^{- 1}$ . KYUN isse ratio ki tarah padho: dB hamesha sirf ratios dekhta hai; percentage ko pehle fraction mein convert karo. $Δ β = 10 lo g_{10} (1 0^{- 1}) = 10 \times (- 1) = - 10 dB .$ Toh 90% energy kaatna sirf 10 dB drop hai — ek warning ki "chhota" dB change kitni energy chupaata hai.

Level 4 — Synthesis

Goal: do independent effects ko ek problem mein saath jodo.

Recall Solution 4.1

KYA karte hain: dono effects ko do intensity ratios ki tarah handle karo aur unhe multiply karo, kyunki dono usi intensity ko scale karte hain. KYUN ratios multiply karo: intensity (sources ki number) ke proportional hai aur $1/ r^{2}$ ke bhi, isliye $I_{final} / I_{ref} = N \times (r_{1} / r_{2})^{2}$ .

Zyaada sources: $\times 4$ .
Door jaana: $(4/12)^{2} = (1/3)^{2} = 1/9$ . $\frac{I _{2}}{I _{1}} = 4 \times \frac{1}{9} = \frac{4}{9} .$ Ab dB mein convert karo: $Δ β = 10 lo g_{10} \frac{4}{9} = 10 (lo g_{10} 4 - lo g_{10} 9) = 10 (0.602 - 0.954) = 10 (- 0.352) = - 3.52 dB .$ $β_{final} = 100 - 3.5 = 96.5 dB (\approx 96 - 97 dB) .$ 4 sirens +6 dB add karte hain, lekin extra distance lagbhag −9.5 dB le jaata hai, net mein ek chhhoti si drop.

Recall Solution 4.2

KYA karte hain: har level ko intensity ratio mein convert karo ( $I_{0}$ ke relative), intensities add karo, wapas convert karo. KYUN: tum 70 aur 74 dB seedha add nahi kar sakte — sirf physical intensities add hoti hain.

$I_{0}$ ki units mein kaam karo taaki hum $1 0^{- 12}$ kabhi touch na karein: $\frac{I _{A}}{I _{0}} = 1 0^{70/10} = 1 0^{7}, \frac{I _{B}}{I _{0}} = 1 0^{74/10} = 1 0^{7.4} .$ Add karo: $\frac{I _{A} + I _{B}}{I _{0}} = 1 0^{7} + 1 0^{7.4} = 1 0^{7} (1 + 1 0^{0.4}) = 1 0^{7} (1 + 2.512) = 1 0^{7} \times 3.512.$ Total ko wapas dB mein convert karo: $β = 10 lo g_{10} (1 0^{7} \times 3.512) = 10 (7 + lo g_{10} 3.512) = 10 (7 + 0.5457) = 75.5 dB .$ Note karo yeh louder source (74 dB) se bilkul thoda upar hai, exactly jaisa expect tha: thoda dheema sound add karna total ko sirf thoda sa nudge karta hai.

Level 5 — Mastery

Goal: pressure–vs–intensity subtlety, inversion, aur ek Doppler-flavoured mix.

Recall Solution 5.1

KYA karte hain: pressure factor ko pehle intensity factor mein convert karo, phir dB mein. KYUN $I \propto p^{2}$ : sound wave ki energy density pressure amplitude ke square ke proportional hoti hai (jaise kinetic energy $\propto v^{2}$ ). Toh $p$ ko 10 se multiply karna $I$ ko $1 0^{2} = 100$ se multiply karta hai. $\frac{I _{2}}{I _{1}} = (\frac{p _{2}}{p _{1}})^{2} = 1 0^{2} = 100.$ $Δ β = 10 lo g_{10} (100) = 10 \times 2 = + 20 dB .$ Equivalently, $Δ β = 20 lo g_{10} (p_{2} / p_{1}) = 20 lo g_{10} 10 = 20$ dB — isliye pressure ratios mein 10 ki jagah 20 ka factor hota hai.

Recall Solution 5.2

KYA karte hain: current level nikalo, 80 dB se compare karo, phir koi bhi excess wapas intensity factor mein convert karo. KYUN dono directions: hum intensity → dB jaate hain judge karne ke liye, phir dB → ratio jaate hain fix prescribe karne ke liye. $β = 10 lo g_{10} \frac{3.2 \times 1 0 ^{- 4}}{1 0 ^{- 12}} = 10 lo g_{10} (3.2 \times 1 0^{8}) = 10 (8 + lo g_{10} 3.2) .$ Kyunki $lo g_{10} 3.2 = lo g_{10} (2^{5} /10) = 5 (0.301) - 1 = 0.505$ : $β = 10 (8.505) = 85.05 dB .$ Yeh 80 dB cap se $5.05 dB$ upar hai — compliant nahi. $5.05$ dB hatane ke liye humen ek intensity factor $f$ chahiye jisme $10 lo g_{10} f = - 5.05$ ho: $lo g_{10} f = - 0.505 \Rightarrow f = 1 0^{- 0.505} = 0.313.$ Unhe intensity ko current value ke lagbhag $31%$ tak reduce karna hoga — yaani ise roughly $3.2$ ke factor se kaatna hoga. (Interesting baat yeh hai ki cut factor wahi $3.2$ hai, kyunki woh exactly $lo g_{10} 3.2$ dB upar the.)

Recall Solution 5.3

KYA karte hain: frequency (jo Doppler shift karta hai) ko intensity (power per area) se alag karo, phir dB change padhlo. KYUN dono independent hain: decibel level sirf $I = P / (4 π r^{2})$ par depend karta hai — radiated power aur distance par. Doppler woh frequency change karta hai jo listener detect karta hai, lekin yeh nahi badalta ki us instant mein har square metre par kitni power cross hoti hai. Pitch aur intensity ek hi wave ke do alag attributes hain; dB scale energy flow measure karta hai, pitch nahi.

Agar radiated power $P$ unchanged hai aur distance $r$ momentarily fixed hai, toh $I$ unchanged hai, isliye intensity ratio exactly $1$ hai: $Δ β = 10 lo g_{10} (1) = 0 dB .$
Toh real life mein approaching train zyaada loud kyun lagti hai? Do reasons, jinmein se koi bhi Doppler pitch shift nahi hai:
1. Distance $r$ actually shrink ho raha hai jab train paas aati hai, aur $I \propto 1/ r^{2}$ isse genuinely zyaada intense banata hai — ek pure distance effect (dekho Problem 2.2 / 3.1).
2. Ek psychoacoustic bias: equal intensity par, zyaada pitch wale tones human ear ko louder feel ho sakte hain (dekho Loudness vs intensity — psychoacoustics). Yeh perception hai, wave ki physics nahi. Doppler frequency shift, apne aap mein fixed distance aur power par consider kiya jaaye, exactly $0$ dB contribute karta hai.

Recall One-line answer key (sirf check karne ke liye dekho)

1.1 → 30 dB · 1.2 → $1 0^{- 5}$ W/m² · 1.3 → +10 dB 2.1 → 48 dB · 2.2 → 64 dB · 2.3 → 1000× 3.1 → 68 dB · 3.2 → 100 machines · 3.3 → −10 dB 4.1 → ≈96.5 dB · 4.2 → ≈75.5 dB 5.1 → +20 dB · 5.2 → 85.05 dB, cut to ≈31% (factor ≈3.2) · 5.3 → 0 dB from Doppler alone