Question

A certain sound source is increased in sound level by$\mathbf{\text{30.0dB}}$. By what multiple is (a) its intensity increased and (b) its pressure amplitude increased?

Short answer

1. The intensity is increased by a multiple factor of $1\times {10}^3$.
2. The pressure amplitude is increased by a multiple of $32$.

Step-by-step solution

Given data

A source sound level is increased by $30dB$.

Determining the concept

Intensity of the sound level can be characterized as the original intensity sound level and final intensity sound level. It is denoted by${\mathbf{\beta }}_1\mathbf{}$and${\mathbf{\beta }}_2$respectively. Also, pressure amplitude of sound waves can be characterized by taking the square root of the total intensity.

The total intensity is given by,

Final level intensity – Original level intensity = Total Intensity

The expression for the sound level is given by,

$\mathit{\beta }\mathbf{=}\mathbf{(}\mathbf{10}\mathbf{}\mathbf{\text{dB}}\mathbf{)}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\frac{\mathbf{I}}{{\mathbf{I}}_{\mathbf{0}}}\mathbf{}$

Here, $\beta$is sound level and l is the final intensity and$I_0$ is the initial intensity.

(a) Determine the multiple of its increased intensity

To find the increased intensity, consider the original intensity$I_1$and final intensity$I_2$.

Also, the original and final sound level can be given as,

$$\begin{gathered} {\beta }_1=10\left(\text{dB}\right)\text{log}\left(\frac{I_1}{I_o}\right) \\ {\beta }_2=10\left(\text{dB}\right)\text{log}\left(\frac{I_2}{I_o}\right) \end{gathered}$$

Since,

${\beta }_2={\beta }_1+30.\text{0dB}$

Therefore, the above equation yields,

$10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)=10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)+30.0\text{dB}$

i.e.,$10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)-10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)=30.0\text{dB}$

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,

$log\left(\frac{I_2}{I_1}\right)=3\text{dB}$

Take each side as a exponent of 10.

Therefore,

$$\begin{gathered} {10}^{log\left(\frac{I_2}{I_1}\right)}=\left(\frac{I_2}{I_1}\right) \\ \left(\frac{I_2}{I_1}\right)={10}^3 \end{gathered}$$

Therefore, the intensity is increased by a multiple factor of $1\times {10}^3$.

(b) Determine the multiple factor by which it’s increased pressure amplitude

To find the pressure amplitude, it is the square root of the intensity. Therefore, it is increased by the factor is,

$\sqrt{1000}=31.62\approx 32$

Hence, the pressure amplitude is increased by a multiple of 32.