Two sounds differ in sound level by 8.3 dB

Two sounds differ in sound level by 8.3 dB. What is the ratio of the greater intensity to the smaller intensity?

The correct answer and explanation is:

Correct Answer: The ratio of the greater intensity to the smaller intensity is approximately 6.76.

Explanation:

To determine the ratio of intensities between two sounds that differ by a given number of decibels (dB), the following formula is used: Sound level difference (dB)=10⋅log⁡10(I2I1)\text{Sound level difference (dB)} = 10 \cdot \log_{10} \left( \frac{I_2}{I_1} \right)

Where:

• I2I_2 is the greater sound intensity
• I1I_1 is the smaller sound intensity
• I2I1\frac{I_2}{I_1} is the ratio we are trying to find

We are given that the sound level difference is 8.3 dB. Substituting into the formula: 8.3=10⋅log⁡10(I2I1)8.3 = 10 \cdot \log_{10} \left( \frac{I_2}{I_1} \right)

Divide both sides by 10: 8.310=log⁡10(I2I1)\frac{8.3}{10} = \log_{10} \left( \frac{I_2}{I_1} \right) 0.83=log⁡10(I2I1)0.83 = \log_{10} \left( \frac{I_2}{I_1} \right)

Now solve by using the inverse log (base 10): I2I1=100.83\frac{I_2}{I_1} = 10^{0.83} I2I1≈6.76\frac{I_2}{I_1} \approx 6.76

This means the more intense sound has approximately 6.76 times the intensity of the less intense sound.

The decibel scale is logarithmic, not linear. This means a small change in decibel level represents a large change in actual sound intensity. Each 10 dB increase represents a tenfold increase in intensity. Therefore, an 8.3 dB difference is slightly less than a tenfold increase, which matches the calculated ratio of 6.76. This concept is critical in fields such as acoustics, audio engineering, and environmental noise control, where sound intensity needs to be accurately quantified and compared.