The sound intensity level at a concert increases from 85 dB to 94 dB when the concert begins - Leaving Cert Physics - Question d - 2009

To determine the factor by which the sound intensity has increased, we first calculate the change in the sound intensity level:

Change in intensity level = 94 dB - 85 dB = 9 dB.

Since an increase of 3 dB occurs when the intensity doubles, we can express the relationship between intensity level and sound intensity mathematically. If we let the factor be represented as $I$, where $I$ is the intensity, the formula related to dB is given by:

$L = 10 imes ext{log}_{10}(I)$

Rearranging, we can see that for an increase of 9 dB, we can equate it as:

rac{L_2 - L_1}{10} = ext{log}_{10}(rac{I_2}{I_1})

From our change in dB (9 dB):

9 = 10 imes ext{log}_{10}(rac{I_2}{I_1})

Dividing both sides by 10 gives us: ext{log}_{10}(rac{I_2}{I_1}) = 0.9

Now, converting from logarithmic form, we exponentiate both sides:

rac{I_2}{I_1} = 10^{0.9} \\ = 7.94.