Q24E - University Physics with Modern Physics

Question

(a) If two sounds differ by 5.00 dB, find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Step-by-Step Solution

Answer

a) The ratio of the intensity is 3.16 b) $\mathrm{\Delta \beta }=20\mathrm{db}$   c) $\mathrm{\Delta \beta }=3\mathrm{db}$ 

STEP 1 T he difference between the sound intensity levels for two different intensities

The formula is given by ${\mathbf{\beta }}_{\mathbf{2}}\mathbf{-}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{log}\mathbf{\left(}\frac{{\mathbf{I}}_{\mathbf{2}}}{{\mathbf{I}}_{\mathbf{1}}}\mathbf{\right)}$ where, ${\mathbf{l}}_{\mathbf{2}}$ is the new intensity and ${\mathbf{l}}_{\mathbf{1}}$ is the previous intensity

STEP 2 Find the ratio of these two intensities

Substitute the values in the equation ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1=10\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\right)$ we get,

$$\begin{gathered} 5\mathrm{db}=10\log \left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\right) \\ 0.5=\log \left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\right) \\ {10}^{0.5}=\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1} \\ \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=3.16 \end{gathered}$$

Step 3 Find the difference in sound intensity level between the two sounds

If the intensity of the first sound is ${\mathrm{l}}_1$, then the intensity of the second sound will be ${\mathrm{l}}_2$ =$100{\mathrm{l}}_1$ , to find the difference in sound intensity level between the two sounds $∆\beta$, we need to substitute into (l) for ${\mathrm{l}}_1$ and ${\mathrm{l}}_2$

$$\begin{gathered} ∆\mathrm{\beta }=10\log \left(\frac{100{\mathrm{l}}_1}{{\mathrm{l}}_1}\right) \\ =10\log \left(100\right) \\ =20\mathrm{db} \end{gathered}$$

 Now the intensity of the second sound will be ${\mathrm{l}}_2=2{\mathrm{l}}_1$. Therefore, we get,

$$\begin{gathered} ∆\mathrm{\beta }=10\log \left(\frac{2{\mathrm{l}}_1}{{\mathrm{l}}_1}\right) \\ =10\log \left(2\right) \\ =3\mathrm{db} \end{gathered}$$

Thus, the difference in sound intensity level between the two sounds is 3db