Two identical speakers each producing 70 dB are placed side by side and both turned on. What is the resulting sound level?
A140 dB — doubling the number of sources doubles the decibel level
B73 dB — doubling the intensity adds approximately 3 dB
C70 dB — the two identical sources cancel each other out
D80 dB — adding a second source always adds 10 dB
Doubling intensity (I₂ = 2I₁) adds 10 log₁₀(2) ≈ 3 dB — not 70 dB. The most tempting wrong answer (140 dB) treats decibels as if they add arithmetically, but the dB scale is logarithmic: it compresses intensity ratios, so equal dB steps represent equal intensity *ratios*, not equal additions. Two identical sources together double the physical intensity, which corresponds to a 3 dB increase.
Question 2 Multiple Choice
A sound source measures 80 dB at 10 meters. You move to 40 meters away. What is the approximate sound level at the new distance?
A20 dB — the level decreases proportionally with distance
B74 dB — the level drops by 6 dB for every factor-of-4 increase in distance
C68 dB — quadrupling the distance drops intensity by a factor of 16, reducing the level by about 12 dB
D60 dB — moving twice as far reduces sound level by 20 dB
Moving from 10 m to 40 m quadruples the distance (×4), which equals two consecutive doublings. The inverse-square law says intensity drops by 4² = 16. Level change = –10 log₁₀(16) ≈ –12 dB, giving 80 – 12 = 68 dB. Equivalently: each doubling of distance subtracts ~6 dB, and two doublings (10→20→40 m) subtract ~12 dB. Option B incorrectly applies the 6 dB rule to a factor-of-4 distance change rather than a factor-of-2.
Question 3 True / False
A sound measuring 60 dB is 10 times more intense than a sound measuring 50 dB.
TTrue
FFalse
Answer: True
This follows directly from the decibel definition: each 10 dB increase corresponds to a factor of 10 in intensity. From β = 10 log₁₀(I/I₀), a difference of 10 dB means 10 = 10 log₁₀(I₂/I₁), so I₂/I₁ = 10. Students sometimes expect a smaller intensity ratio for a modest-sounding 10 dB difference, but the logarithmic scale is precisely designed so that equal dB intervals represent equal intensity ratios.
Question 4 True / False
A 40 dB sound is twice as intense as a 20 dB sound.
TTrue
FFalse
Answer: False
40 dB and 20 dB differ by 20 dB, which corresponds to a factor of 10²⁰/¹⁰ = 100 in intensity — not a factor of 2. 'Twice as intense' would be an increase of only 10 log₁₀(2) ≈ 3 dB. This is one of the most common misconceptions about the decibel scale: the numbers feel linear (40 is twice 20), but they represent a logarithmic relationship where the intensity ratio is 100, not 2.
Question 5 Short Answer
Why does the decibel scale use a logarithm rather than expressing sound intensity directly in watts per square meter?
Think about your answer, then reveal below.
Model answer: The human auditory system spans an intensity range of roughly 10¹² — from the threshold of hearing at 10⁻¹² W/m² to the pain threshold near 1 W/m². This makes direct W/m² values inconveniently small and hard to compare. More fundamentally, the ear responds roughly logarithmically: equal ratios of intensity produce equal perceived differences in loudness. The decibel logarithm converts multiplicative intensity relationships into additive dB steps, compressing a trillion-to-one physical range into a 0–120 dB scale that tracks how loudness is actually perceived.
The logarithm is not just computational convenience — it reflects the structure of human perception. The ear is a ratio detector, not a difference detector. This is why both the decibel scale and the musical frequency scale (octaves) are logarithmic: they match the perceptual architecture of the auditory system.