When the intensity doubles, the relative change is:

When the intensity doubles, the relative change is:

• A. 6 dB
• B. 3 dB
• C. -3 dB
• D. 10 dB

Answer: (B)

Decibels express ratios on a logarithmic scale, using dB = 10 log10(I2/I1) for intensity. If the intensity doubles, I2/I1 = 2, so the change is 10 log10(2) ≈ 3.01 dB, which rounds to 3 dB. So the relative change when intensity doubles is about +3 dB. For context: 4x intensity would yield about +6 dB, 10x would be +10 dB, and halving the intensity would be about -3 dB. The problem uses intensity, not pressure, so doubling the intensity gives the +3 dB result.

Decibels express ratios on a logarithmic scale, using dB = 10 log10(I2/I1) for intensity. If the intensity doubles, I2/I1 = 2, so the change is 10 log10(2) ≈ 3.01 dB, which rounds to 3 dB. So the relative change when intensity doubles is about +3 dB.

For context: 4x intensity would yield about +6 dB, 10x would be +10 dB, and halving the intensity would be about -3 dB. The problem uses intensity, not pressure, so doubling the intensity gives the +3 dB result.