Questions: Sound Intensity Level and the Decibel Scale

Question 1

Two identical loudspeakers, each producing 80 dB, are placed side by side and play simultaneously. The combined sound level is approximately:

• A. 160 dB — you add the decibel values of both sources
• B. 80 dB — the two sources are in phase and cancel
• C. 83 dB — doubling intensity adds approximately 3 dB
• D. 90 dB — each additional identical source adds 10 dB

Answer: (C)

80 dB corresponds to an intensity of I₀ × 10⁸. Two identical sources together produce 2 × I₀ × 10⁸. Applying the formula: L = 10 log₁₀(2 × 10⁸) = 10(log₁₀2 + 8) ≈ 10(0.3 + 8) = 83 dB. Decibels are logarithmic — you cannot add them linearly. The common error of adding the dB values (80 + 80 = 160) confuses the compressed logarithmic representation with the underlying intensity values.

Question 2

A sound level increases from 60 dB to 90 dB. By what factor has the intensity increased?

• A. 1.5 times — the ratio 90/60
• B. 30 times — the difference 90 − 60
• C. 1000 times — 10^(30/10) = 10³
• D. 2 times — one additional octave of loudness

Answer: (C)

A difference of ΔL dB corresponds to an intensity ratio of 10^(ΔL/10). For ΔL = 30 dB: 10^(30/10) = 10³ = 1000. The tempting wrong answer is 30 (the arithmetic difference), which confuses the dB value with the intensity ratio. 'Adding 30 dB' means multiplying intensity by 1000, not adding 30 units of intensity.

Question 3

A 20 dB increase in sound level corresponds to a 100-fold increase in intensity.

• T. True
• F. False

Answer: True

Using the formula: ΔL = 10 log₁₀(I₂/I₁). For ΔL = 20: 20 = 10 log₁₀(ratio), so log₁₀(ratio) = 2, giving ratio = 10² = 100. Each 10 dB increase multiplies intensity by 10, so two successive 10 dB increases multiply by 10 × 10 = 100. The logarithmic scale converts multiplicative factors into additive increments.

Question 4

Ten identical jackhammers produce a sound level approximately ten times higher in decibels than a single jackhammer.

• T. True
• F. False

Answer: False

Ten identical sources produce 10 times the intensity of one, which corresponds to an increase of 10 log₁₀(10) = 10 dB — not 10 times the decibel value. If one jackhammer produces 100 dB, ten together produce about 110 dB, not 1000 dB. The confusion arises from treating decibel values as if they were linear quantities that scale with the number of sources. The logarithmic scale means that adding more sources has diminishing returns in perceived loudness.

Question 5

Why do two identical 70 dB sound sources together produce approximately 73 dB rather than 140 dB?

Model answer: Because decibels are a logarithmic scale encoding intensity *ratios*, not absolute intensities. Each 70 dB source has intensity I = I₀ × 10⁷. Two together produce 2I₀ × 10⁷. Converting: L = 10 log₁₀(2 × 10⁷) = 10(log₁₀2 + 7) ≈ 10(0.301 + 7) = 73 dB. Adding the sources doubles the intensity, and doubling intensity corresponds to +3 dB regardless of the starting level. The number 140 dB would require an intensity 10¹⁴ times the reference — a level that would destroy hearing instantly, which is clearly wrong for two ordinary speakers.

The 140 dB error comes from treating dB like a count. But dB encodes a ratio logarithmically. You must convert to intensity, add intensities, then convert back. The practical lesson: combining identical sources never more than doubles intensity, adding at most 3 dB per doubling of the number of sources — so 10 sources add only 10 dB, and 100 sources add only 20 dB.