Question

If the intensity of sound is doubled, by how many decibels does the sound level increase?

Answer 1

Let the intensity of the sound be I and ${\beta }_1$ be the sound level. If the intensity of the sound is doubled, then its sound level becomes 2I.
Sound level ${\beta }_1$ is given by:
${\beta }_1=10{\log }_{10}\frac{I}{I_0}$,
where I₀ is the constant reference intensity.
When the intensity doubles, the sound level is given by:
${\beta }_2=10{\log }_{10}\frac{2I}{I_0}$.
According to the question,
$$\begin{gathered} {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10\log \left(\frac{2I}{I}\right) \\ =10\times 0.3010=3\mathrm{dB} \end{gathered}$$

The sound level is increased by 3 dB.