Question

At a distance of 5 m from a point sound source, the sound intensity level is 110 dB. At what distance is the intensity level 95dB?

Answer 1

The intencity of the sound produced by a point source is inversely proportional to the squared distance from this source. Thus, obtain:

$$I_1\propto\dfrac{1}{r_1^2}\\ I_2\propto\dfrac{1}{r_2^2}$$

where $I_1$ and $I_2$ are intencities at the distances $r_1 = 5m$ and $r_2$ respectively. Thus, obtain:

$$r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}$$

By definition, dB are:

$$110dB = 10\lg I_1\\ 90dB = 10\lg I_2$$

From this obtain:

$$I_1 = 10^{11}\space (relative\space units)\\ I_2 = 10^{9}\space (relativ\space units)$$

Exact units don't matter since we are only interested in ratio $I_1/I_2$. Substituting into the expression for $r_2$, obtain:

$$r_2 = 5\sqrt{\dfrac{10^{11}}{10^9}} = 50m$$

Answer. 50m.