Question

Two sounds differ in sound level by$\mathbf{\text{1.00dB}}$. What is the ratio of the greater intensity to the smaller intensity?

Short answer

The ratio of greater intensity and smaller intensity is$1.26$ .

Step-by-step solution

Given

Change in sound level,$\text{Δ}\beta =1.0\text{dB}$

Determining the concept

The sound level$\mathbf{\beta }\mathbf{}$in decibels (dB) can be defined as,

$\mathbf{\beta }\mathbf{=}\text{10}\left(\text{dB}\right)\text{log}(\mathbf{I}\mathbf{/}{\mathbf{I}}_0)$

where,${\mathbf{I}}_0\mathbf{}$is the reference intensity level to which all intensities are compared. For every multiple of 10 of intensity, 10 dB is added to the sound level.

Determining the

To find the ratio of the greater intensity to the lower intensity, consider the original intensity$I_1$and final intensity$I_2$. Also, the original and final sound level can be given as,

$$\begin{gathered} {\beta }_1=10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right) \\ {\beta }_2=10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right) \\ \begin{array}{c}\text{Δ}\beta ={\beta }_2-{\beta }_1\\ =1.\text{0dB}\end{array} \end{gathered}$$

Thus, the above equation becomes,

$\begin{array}{l}10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)=10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)+1.0\text{dB}\\ 10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)-10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)=1.0\text{dB}\end{array}$

Dividing the above equation by 10 dB and using identity,

$\log \left(\frac{I_2}{I_0}\right)-\log \left(\frac{I_1}{I_0}\right)=\log \left(\frac{I_2}{I_1}\right)$

Therefore,

$\log \left(\frac{I_2}{I_1}\right)=0.1\text{dB}$

Now, using each side as exponent of 10.Therefore,

$$\begin{gathered} {10}^{\log \left(\frac{I_2}{I_1}\right)}=\left(\frac{I_2}{I_1}\right) \\ \left(\frac{I_2}{I_1}\right)={10}^{0.1} \\ \left(\frac{I_2}{I_1}\right)=1.26 \end{gathered}$$

Therefore, the ratio of greater intensity and smaller intensity is$1.26$