Question

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

Answer 1

(a) Let $I_1$ be the original intensity and $I_2$ be the final intensity.

The original sound level is ${\beta }_1=\left(10dB\right)\log \left(I_1/I_0\right)$

and the final level ${\beta }_2=\left(10dB\right)\log \left(I_2/I_0\right)$,

where $I_0$ is the reference intensity.

Since ${\beta }_2={\beta }_1+30dB$, which yields

$\left(10dB\right)\log \left(I_2/I_0\right)=\left(10dB\right)\log \left(I_1/I_0\right)+30dB$,
or
$\left(10dB\right)\log \left(I_2/I_0\right)-\left(10dB\right)\log \left(I_1/I_0\right)=30dB$

Divided by 10 dB and use $\log \left(I_2/I_0\right)-\log \left(I_1/I_0\right)=\log \left(I_2/I_1\right)$ to obtain $\log \left(I_2/I_1\right)=3$. Now use each side as an exponent of 10 and recognize that ${10}^{\log \left(I_2/I_1\right)}=I_2/I_1$. The result is $I_2/i_1={10}^3$. The intensity is increased by a factor of $1.0\times {10}^3$.

(b) The pressure amplitude is proportional to the square root of the intensity, so it is increased by a factor of $\sqrt{1000}\approx 32$.