Intuition The big picture (WHY decibels exist)Your ear is insanely sensitive. The quietest sound you can hear has intensity ∼ 10 − 12 W/m 2 \sim 10^{-12}\ \text{W/m}^2 ∼ 1 0 − 12 W/m 2 ∼ 1 W/m 2 \sim 1\ \text{W/m}^2 ∼ 1 W/m 2 a trillion times . Writing 0.0000000000034 W/m 2 0.0000000000034\ \text{W/m}^2 0.0000000000034 W/m 2 compress this huge range with a logarithm. A log turns "multiply by 10" into "add 1". The decibel scale is just "how many factors of 10 louder than the quietest sound am I?", scaled nicely. Logarithms turn multiplication into addition — that's the entire trick.
Definition Sound intensityI I I Intensity is the power carried by a sound wave per unit area facing the wave:
I = P A [ W/m 2 ] I = \frac{P}{A} \qquad [\text{W/m}^2] I = A P [ W/m 2 ] P P P A A A
WHY this definition? Sound carries energy. Spread that energy over a big area and each square metre gets a small share (quiet); concentrate it on a small area and each metre gets more (loud). Intensity measures energy flow density .
Intuition Why intensity falls off as1 / r 2 1/r^2 1/ r 2 r r r A = 4 π r 2 A = 4\pi r^2 A = 4 π r 2 I = P 4 π r 2 ∝ 1 r 2 . I = \frac{P}{4\pi r^2}\propto \frac{1}{r^2}. I = 4 π r 2 P ∝ r 2 1 . quarter . Nothing is "lost"; it's just spread thinner.
We want a number that:
grows by a fixed step every time intensity ×10,
starts at 0 for the threshold of hearing.
Definition Reference intensityThe agreed quietest audible intensity (the threshold of hearing at 1 kHz):
I 0 = 10 − 12 W/m 2 . I_0 = 10^{-12}\ \text{W/m}^2. I 0 = 1 0 − 12 W/m 2 .
Step 1 — the bel. Ask "how many factors of 10 is I I I I 0 I_0 I 0 β bel = log 10 ( I I 0 ) . \beta_{\text{bel}} = \log_{10}\!\left(\frac{I}{I_0}\right). β bel = log 10 ( I 0 I ) . Why log? Because log ( I / I 0 ) \log(I/I_0) log ( I / I 0 ) I = 1000 I 0 I = 1000\,I_0 I = 1000 I 0 log 10 ( 1000 ) = 3 \log_{10}(1000)=3 log 10 ( 1000 ) = 3
Step 2 — the deci. One bel is a coarse step (a whole factor of 10). We want finer resolution, so we use tenths of a bel = decibels. Multiply by 10:
HOW to read it: β \beta β level (a comparison to I 0 I_0 I 0 not an intensity. It is dimensionless but we tack on "dB" to remember the convention.
Suppose intensity becomes 10 I 10I 10 I β ′ = 10 log 10 10 I I 0 = 10 [ log 10 10 + log 10 I I 0 ] = 10 ( 1 ) + β = β + 10. \beta' = 10\log_{10}\!\frac{10I}{I_0} = 10\Big[\log_{10}10 + \log_{10}\tfrac{I}{I_0}\Big] = 10\big(1\big) + \beta = \beta + 10. β ′ = 10 log 10 I 0 10 I = 10 [ log 10 10 + log 10 I 0 I ] = 10 ( 1 ) + β = β + 10.
Worked example (a) Intensity → dBA jackhammer gives I = 10 − 2 W/m 2 I = 10^{-2}\ \text{W/m}^2 I = 1 0 − 2 W/m 2 β \beta β β = 10 log 10 10 − 2 10 − 12 = 10 log 10 ( 10 10 ) = 10 × 10 = 100 dB . \beta = 10\log_{10}\!\frac{10^{-2}}{10^{-12}} = 10\log_{10}(10^{10}) = 10\times 10 = 100\ \text{dB}. β = 10 log 10 1 0 − 12 1 0 − 2 = 10 log 10 ( 1 0 10 ) = 10 × 10 = 100 dB . Why this step? 10 − 2 10 − 12 = 10 10 \frac{10^{-2}}{10^{-12}} = 10^{10} 1 0 − 12 1 0 − 2 = 1 0 10 log 10 ( 10 10 ) = 10 \log_{10}(10^{10})=10 log 10 ( 1 0 10 ) = 10
Worked example (b) dB → intensity (invert the log)A normal conversation is 60 dB 60\ \text{dB} 60 dB I I I 60 = 10 log 10 I I 0 ⇒ log 10 I I 0 = 6 ⇒ I I 0 = 10 6 . 60 = 10\log_{10}\frac{I}{I_0}\ \Rightarrow\ \log_{10}\frac{I}{I_0}=6\ \Rightarrow\ \frac{I}{I_0}=10^{6}. 60 = 10 log 10 I 0 I ⇒ log 10 I 0 I = 6 ⇒ I 0 I = 1 0 6 . I = 10 6 × 10 − 12 = 10 − 6 W/m 2 . I = 10^{6}\times 10^{-12} = 10^{-6}\ \text{W/m}^2. I = 1 0 6 × 1 0 − 12 = 1 0 − 6 W/m 2 . Why this step? Divide by 10 to undo the ×10, then raise 10 to both sides to undo log 10 \log_{10} log 10
Worked example (c) Adding identical sourcesOne violin is 50 dB 50\ \text{dB} 50 dB ten identical violins together?
Ten sources → 10× the intensity (intensities add).
Δ β = 10 log 10 ( 10 ) = + 10 dB ⇒ 60 dB . \Delta\beta = 10\log_{10}(10) = +10\ \text{dB}\ \Rightarrow\ 60\ \text{dB}. Δ β = 10 log 10 ( 10 ) = + 10 dB ⇒ 60 dB . Why this step? We compare new vs old intensity; the I 0 I_0 I 0 ratio 10 10 10
Worked example (d) Distance & inverse-square togetherAt 2 m 2\ \text{m} 2 m 80 dB 80\ \text{dB} 80 dB 8 m 8\ \text{m} 8 m 4 2 = 16 4^2 = 16 4 2 = 16 Δ β = 10 log 10 1 16 = − 10 log 10 16 = − 10 ( 1.204 ) = − 12.0 dB . \Delta\beta = 10\log_{10}\!\frac{1}{16} = -10\log_{10}16 = -10(1.204) = -12.0\ \text{dB}. Δ β = 10 log 10 16 1 = − 10 log 10 16 = − 10 ( 1.204 ) = − 12.0 dB . = 80 − 12 = 68 dB = 80 - 12 = 68\ \text{dB} = 80 − 12 = 68 dB Why this step? I ∝ 1 / r 2 I\propto 1/r^2 I ∝ 1/ r 2 I 2 / I 1 = ( r 1 / r 2 ) 2 = ( 2 / 8 ) 2 = 1 / 16 I_2/I_1 = (r_1/r_2)^2 = (2/8)^2 = 1/16 I 2 / I 1 = ( r 1 / r 2 ) 2 = ( 2/8 ) 2 = 1/16
Common mistake "Two 60 dB speakers = 120 dB."Why it feels right: adding things normally means adding their numbers, and dB are numbers.
The fix: dB are logarithmic — you can't add levels directly. Add intensities : two equal sources double the intensity → + 10 log 10 2 ≈ + 3 dB +10\log_{10}2 \approx +3\ \text{dB} + 10 log 10 2 ≈ + 3 dB 63 dB , not 120 dB.
Common mistake "120 dB is twice as loud/intense as 60 dB."Why it feels right: 120 = 2 × 60 120 = 2\times 60 120 = 2 × 60 The fix: The difference is 120 − 60 = 60 dB = 6 120-60 = 60\ \text{dB} = 6 120 − 60 = 60 dB = 6 I I I 10 6 = 10^6 = 1 0 6 = a million times larger. dB measure ratios , never simple multiples.
20 log 10 20\log_{10} 20 log 10 Why it feels right: you've seen 20 log 10 20\log_{10} 20 log 10 The fix: 20 log 10 20\log_{10} 20 log 10 amplitude/pressure ratios (because I ∝ p 2 I\propto p^2 I ∝ p 2 log I = 2 log p \log I = 2\log p log I = 2 log p intensity , it's 10 log 10 10\log_{10} 10 log 10
Common mistake ForgettingI 0 I_0 I 0 I I I I 0 I_0 I 0 Why it feels right: the log "hides" units.
The fix: I / I 0 I/I_0 I / I 0 W/m 2 \text{W/m}^2 W/m 2
Source
I I I β \beta β
Threshold of hearing
10 − 12 10^{-12} 1 0 − 12 0
Whisper
10 − 10 10^{-10} 1 0 − 10 20
Conversation
10 − 6 10^{-6} 1 0 − 6 60
Busy street
10 − 4 10^{-4} 1 0 − 4 80
Jackhammer
10 − 2 10^{-2} 1 0 − 2 100
Pain / damage
1 1 1 120
Notice: each row jumps by powers of ten in I I I That's the compression.
Recall Feynman: explain to a 12-year-oldImagine sound as rain hitting your hand. A drizzle is "1 drop", a downpour is "a million drops". If you wrote those numbers on the same chart, the drizzle would be invisible. So instead of writing the number of drops, you write how many zeros it has. The decibel is basically "count the zeros, then ×10". 0 dB = barely-there drizzle, 120 dB = wall of rain that hurts. Adding 10 to the dB number means ten times more rain , not just a little more.
Mnemonic Remember the constants & the +10 rule"Twelve-tiny, ten-times-ten."
I 0 = 10 − 12 I_0 = 10^{-12} I 0 = 1 0 − 12 twelve zeros tiny.Formula has ten out front: β = 10 log 10 ( I / I 0 ) \beta = \mathbf{10}\log_{10}(I/I_0) β = 10 log 10 ( I / I 0 )
×10 → +10 dB ; ×2 → +3 dB ("two is three").
What is the SI unit of sound intensity? watts per square metre,
W/m 2 \text{W/m}^2 W/m 2 Define sound intensity. Power per unit area carried by the wave,
I = P / A I = P/A I = P / A .
Write the decibel level formula. β = 10 log 10 ( I / I 0 ) \beta = 10\log_{10}(I/I_0) β = 10 log 10 ( I / I 0 ) with
I 0 = 10 − 12 W/m 2 I_0 = 10^{-12}\,\text{W/m}^2 I 0 = 1 0 − 12 W/m 2 .
What is the reference intensity I 0 I_0 I 0 10 − 12 W/m 2 10^{-12}\,\text{W/m}^2 1 0 − 12 W/m 2 ; the threshold of human hearing at 1 kHz.
By how many dB does the level change if intensity is multiplied by 10? +10 dB.
By how many dB does the level change if intensity is doubled? about +3 dB (
10 log 10 2 10\log_{10}2 10 log 10 2 ).
Why use a logarithmic scale for sound? The audible intensity range spans ~
10 12 10^{12} 1 0 12 , so a log compresses it into a manageable 0–120 scale and matches roughly how loudness is perceived.
Two equal sources at 60 dB combine to what level? 63 dB (intensities add → ×2 → +3 dB), NOT 120 dB.
How does intensity vary with distance from a point source? I = P / ( 4 π r 2 ) ∝ 1 / r 2 I = P/(4\pi r^2)\propto 1/r^2 I = P / ( 4 π r 2 ) ∝ 1/ r 2 .
A sound reads 80 dB at 2 m; what at 8 m (free field)? r r r ×4 →
I I I ÷16 →
− 10 log 10 16 ≈ − 12 -10\log_{10}16\approx-12 − 10 log 10 16 ≈ − 12 dB → 68 dB.
Convert 60 dB to intensity. I = I 0 10 6 = 10 − 6 W/m 2 I = I_0\,10^{6} = 10^{-6}\,\text{W/m}^2 I = I 0 1 0 6 = 1 0 − 6 W/m 2 .
Why is it 10 log 10 10\log_{10} 10 log 10 20 log 10 20\log_{10} 20 log 10 Because
I ∝ p 2 I\propto p^2 I ∝ p 2 , so
log I = 2 log p \log I = 2\log p log I = 2 log p , doubling the prefactor.
How many times more intense is 120 dB than 60 dB? 10 ( 120 − 60 ) / 10 = 10 6 10^{(120-60)/10}=10^6 1 0 ( 120 − 60 ) /10 = 1 0 6 = one million times.
Derive the difference rule Δ β = 10 log 10 ( I 2 / I 1 ) \Delta\beta = 10\log_{10}(I_2/I_1) Δ β = 10 log 10 ( I 2 / I 1 ) Subtract
β 2 − β 1 \beta_2-\beta_1 β 2 − β 1 ; the
log I 0 \log I_0 log I 0 terms cancel, leaving
10 log 10 ( I 2 / I 1 ) 10\log_{10}(I_2/I_1) 10 log 10 ( I 2 / I 1 ) .
Sound waves — pressure & displacement (intensity ∝ \propto ∝ Inverse-square law for radiation (where 1 / r 2 1/r^2 1/ r 2 Logarithms and exponentials (math engine of the scale)Loudness vs intensity — psychoacoustics (perception ≈ log of stimulus, Weber–Fechner)Wave energy and power (defines the P P P I = P / A I=P/A I = P / A Doppler effect (companion sound-wave phenomenon)
point source spreads on sphere
Use logarithm to compress
Threshold I0 = 10^-12 W/m2
Difference rule delta-beta
Intuition Hinglish mein samjhoDekho, hamare kaan bahut sensitive hote hain — sabse halki sunai dene wali awaaz ki intensity 10 − 12 W/m 2 10^{-12}\ \text{W/m}^2 1 0 − 12 W/m 2 1 W/m 2 1\ \text{W/m}^2 1 W/m 2 ek trillion guna hai! Itni badi range ko seedha likhna impossible hai. Isiliye hum logarithm lagate hain — log ka magic yeh hai ki "×10 karna" ko "+1 karna" bana deta hai. Bas yahi decibel scale ka asli idea hai.
Formula hai β = 10 log 10 ( I / I 0 ) \beta = 10\log_{10}(I/I_0) β = 10 log 10 ( I / I 0 ) I 0 = 10 − 12 W/m 2 I_0 = 10^{-12}\ \text{W/m}^2 I 0 = 1 0 − 12 W/m 2 10 guna badhti hai, dB sirf +10 hota hai. Aur jab intensity double hoti hai, dB sirf +3 hota hai ("two is three" yaad rakho). Yahi reason hai ki do 60 dB speaker milke 120 dB nahi, balki sirf 63 dB dete hain — kyunki intensity add hoti hai, dB directly add nahi hote.
Distance ka bhi dhyan rakho: point source se I ∝ 1 / r 2 I \propto 1/r^2 I ∝ 1/ r 2 4 π r 2 4\pi r^2 4 π r 2 − 10 log 10 4 ≈ − 6 -10\log_{10}4 \approx -6 − 10 log 10 4 ≈ − 6 intensity ka ratio nikalo, phir 10 log 10 ( ratio ) 10\log_{10}(\text{ratio}) 10 log 10 ( ratio )
