Question

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

Answer 1

Step 1: Given data

Let the intensity of the sound be $\mathrm{I}$.

According to the question, $\mathrm{I}\text{'}=2\mathrm{I}$

Step 2: Formula used

Let the intensity of the sound be $\mathrm{I}$ and it's sound level is ${\mathrm{\beta }}_1$

The sound level $\mathrm{\beta }$ is given by, $\mathrm{\beta }=10{\log }_{10}\left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)$

Where the constant reference intensity is ${\mathrm{I}}_0$

Step 3: The sound level increased

The sound level ${\mathrm{\beta }}_1$ is given by, ${\mathrm{\beta }}_1=10{\log }_{10}\left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)...\left(1\right)$

When the intensity doubles, the sound level is given by, ${\mathrm{\beta }}_2=10{\log }_{10}\left(\frac{2\mathrm{I}}{{\mathrm{I}}_0}\right)...\left(2\right)$

By subtracting equations (2) and (1), the sound level increased is,

$$\begin{gathered} {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=\left(10{\log }_{10}\left(\frac{2\mathrm{I}}{{\mathrm{I}}_0}\right)\right)-\left(10{\log }_{10}\left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)\right) \\ {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10\left({\log }_{10}\left(\frac{2\mathrm{I}}{{\mathrm{I}}_0}\right)-{\log }_{10}\left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)\right)...\left(3\right) \end{gathered}$$

Since ${\log }_a\left(x\right)-{\log }_a\left(y\right)={\log }_a\left(\frac{x}{y}\right)$ equation 3 becomes,

$$\begin{gathered} {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10\left({\log }_{10}\left(\frac{2\mathrm{I}}{{\mathrm{I}}_0}\right)-{\log }_{10}\left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)\right) \\ {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10{\log }_{10}\left(\frac{2\mathrm{I}}{\mathrm{I}}\right) \\ {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10{\log }_{10}2 \\ {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10\times 0.3010 \\ {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=3\mathrm{dB} \end{gathered}$$

Thus, the sound level increases by $3\mathrm{dB}$.