Sound intensity — decibels (logarithmic scale): Question bank

1.6.23 · D5 · Physics › Oscillations & Waves › Sound intensity — decibels (logarithmic scale)

True or false — justify

Do 60 dB speakers milke 120 dB produce karte hain.

False. Levels logarithmic hote hain aur directly kabhi add nahi hote; intensities add hoti hain,

I

double hoti hai, jisse sirf

10 lo g_{10} 2 \approx 3

dB add hota hai → 63 dB.

0 dB sound ka matlab hai "bilkul koi sound nahi".

False. 0 dB hearing ka threshold hai

I_{0} = 1 0^{- 12} W/m^{2}

— ek real (barely audible) intensity, zero nahi. Silence

- \infty

dB hoga kyunki

lo g_{10} (0)

diverge karta hai.

Negative decibel levels impossible hote hain.

False.

I_{0}

se koi bhi intensity kam ho toh

β < 0

milta hai; jaise

I = 1 0^{- 13} W/m^{2}

deta hai

β = 10 lo g_{10} (1 0^{- 1}) = - 10

dB. Negative ka matlab sirf "reference se quieter" hai.

Intensity double karne par hamesha same +3 dB add hota hai, chahe kisi bhi level se shuru karo.

True. Difference rule

Δ β = 10 lo g_{10} (I_{2} / I_{1})

sirf ratio pe depend karta hai; 2 ka ratio hamesha

\approx + 3

dB deta hai, chahe 20 dB se shuru karo ya 100 dB se.

Kyunki

I \propto 1/ r^{2}

hai, point source se apni distance double karne par decibel reading half ho jaati hai.

False. Distance double karne se intensity quarter ho jaati hai, aur

10 lo g_{10} (1/4) \approx - 6

dB — ek subtraction hai, dB number ki halving nahi. Dekho Inverse-square law for radiation.

Decibel ek unit hai jaise metre ya watt.

False. dB dimensionless hai — ye ek ratio

I / I_{0}

ko label karta hai. Hum "dB" sirf isliye likhte hain taaki record rahe ki humne kaunsa reference aur kaunsa prefactor (10 vs 20) use kiya.

Agar tumhe pata hai koi sound 40 dB hai toh tum uski intensity exactly compute kar sakte ho.

True, bas ye bhi pata hona chahiye ki reference

I_{0} = 1 0^{- 12} W/m^{2}

hai; tab

I = I_{0} 1 0^{40/10} = 1 0^{- 8} W/m^{2}

. Reference ke bina akela dB number meaningless hai.

100 dB ki sound 10 dB wali sound se ten times intensity deliver kar rahi hai.

False. Difference 90 dB hai

= 9

factors of ten, isliye

I

1 0^{9}

times larger hai, 10 times nahi. dB differences exponents hain, multipliers nahi.

Spot the error

"Das identical drummers har ek 70 dB par milke

10 \times 70 = 700

dB dete hain."

Tumne levels multiply kar diye. Das sources das times intensity dete hain, isliye

Δ β = 10 lo g_{10} 10 = + 10

dB → 80 dB, 700 nahi.

"Problem ne mujhe pressure ratio 100 diya hai, toh level hai

10 lo g_{10} (100) = 20

dB."

Wrong prefactor. Pressure/amplitude ke liye

20 lo g_{10}

use karo kyunki

I \propto p^{2}

; yahan

20 lo g_{10} (100) = 40

dB hai. Dekho Sound waves — pressure & displacement.

I = 3 \times 1 0^{- 8} μ W/m^{2}

aur

I_{0} = 1 0^{- 12} W/m^{2}

, toh

β = 10 lo g_{10} (3 \times 1 0^{- 8} /1 0^{- 12})

Units mismatch hai:

I

microwatts mein hai,

I_{0}

watts mein. Pehle convert karo (

1 μ W = 1 0^{- 6} W

) taaki log lene se pehle ratio ek pure number ho.

"60 dB se 63 dB jaana sirf 3-unit change hai, toh basically intensity mein kuch nahi hua."

3 dB rise ka matlab hai intensity double ho gayi. Chhote dB steps bade physical changes chhupa dete hain — compression hi toh is scale ka poora point hai.

"Kyunki

lo g_{10} (0) = 0

hai, zero intensity wala source 0 dB read karta hai."

lo g_{10} (0)

zero nahi hai — ye

- \infty

hai. Zero intensity ki koi finite decibel value nahi hoti; 0 dB correspond karta hai

I = I_{0}

se, na ki

I = 0

se.

120

dB,

60

dB se twice as intense hai kyunki

120 = 2 \times 60

Scale logarithmic hai, isliye tum divide nahi karte, subtract karte ho:

120 - 60 = 60

= 1 0^{6}

, matlab ek million times zyada intense.

Why questions

Hum intensity ka log kyun lete hain instead of directly intensity

W/m^{2}

mein quote karne ke?

Audible intensities

\sim 1 0^{12}

tak span karti hain; log "multiply by 10" ko "add 1" mein badal deta hai, us trillion-fold range ko 0–120 ke readable scale mein compress karke — dekho Logarithms and exponentials.

Prefactor 10 (deci-bel) kyun hai na ki 1 (bel)?

Ek bel ek coarse step hai (ek full factor of ten). 10 se multiply karne par tenths-of-a-bel resolution milta hai, toh barely-noticeable change roughly 1 dB register hoti hai instead of fraction of a bel ke.

Jab hum do sounds compare karte hain toh reference

I_{0}

kyun disappear ho jaata hai?

β_{2} - β_{1}

subtract karne par milta hai

10 lo g_{10} (I_{2} / I_{0}) - 10 lo g_{10} (I_{1} / I_{0}) = 10 lo g_{10} (I_{2} / I_{1})

;

I_{0}

wale terms cancel ho jaate hain, isliye ratio problems mein kabhi iski zaroorat nahi.

Intensities add hoti hain lekin decibel levels nahi, kyun?

Intensity energy flow per area hai — physically additive. Decibel intensity ka log hai, aur

lo g (a + b) \neq = lo g a + lo g b

, isliye tumhe intensity par wapas jaana hoga, sum karna hoga, phir re-log karna hoga. Dekho Wave energy and power.

Decibel scale cheezein kitni loud lagti hain usse raw intensity se better kyun match karti hai?

Human loudness perception roughly logarithmic hai (Weber–Fechner), isliye equal ratio increases equal steps jaisi feel hoti hain — exactly wahi jo log encode karta hai. Detail mein hai Loudness vs intensity — psychoacoustics mein.

Pressure ke liye

20 lo g_{10}

kyun use hota hai lekin intensity ke liye

10 lo g_{10}

Kyunki intensity pressure squared ke proportional hai,

lo g I = 2 lo g p

;

10 lo g_{10} (I / I_{0})

mein substitute karne par 2 bahar aa jaata hai, deta hai

20 lo g_{10} (p / p_{0})

Edge cases

I = I_{0}

exactly ke liye kaunsa decibel level correspond karta hai?

β = 10 lo g_{10} (1) = 0

dB — definition anchor. Isliye hearing ka threshold natural zero hai, silence nahi.

Jab

I \to 0

hota hai toh

β

ka kya hota hai?

β \to - \infty

, kyunki zero ki taraf jaate number ka

lo g_{10}

unboundedly negative hota jaata hai. Real silence ki koi finite decibel reading nahi hoti.

Agar ek point source ki power

P

double ho jaaye lekin tum bhi twice dur chale jao, toh net dB change kya hai?

Intensity scale karti hai

P / r^{2}

ke according: power ×2 deta hai

+ 3

dB, distance ×2 deta hai

- 6

dB, net

- 3

dB — ratios dB mein additively combine hote hain.

Kya 50 dB par do alag-frequency wale tones ko intensities mein same tarah add kiya ja sakta hai jaise identical sources ko?

Haan total energy ke liye: uncorrelated intensities add hoti hain, deta hai

+ 3

dB → 53 dB. (Sirf coherent same-phase waves interfere kar sakti hain; independent tones nahi karte.)

Ek moving siren apni pitch Doppler effect ke through shift karti dikhti hai — kya wo Doppler shift khud uska decibel level change karta hai?

Nahi. Doppler observed frequency change karta hai, intensity nahi, isliye pitch shift se level unchanged rehta hai; level sirf accompanying distance (inverse-square) variation se change hota hai.

Agar tum ek free field mein point source ki distance half kar do, toh exact dB change kya hoga?

Distance ÷2 → intensity ×4 →

Δ β = 10 lo g_{10} 4 \approx + 6

dB. Distance ki har halving approximately 6 dB add karti hai.

Recall One-line trap detector

Koi bhi dB question answer karne se pehle, do cheezein poochho ::: (1) Kya ye ek absolute level hai (needs $I_{0}$ ) ya ek difference hai (needs only the ratio)? (2) Kya di gayi quantity intensity hai ( $10 lo g$ ) ya pressure/amplitude ( $20 lo g$ )?