Question

The sound level measured in a room by a person watching a movie on a home theater system varies from 60 dB during a quiet part to 100 dB during a loud part. Approximately how many times louder is the latter sound

Answer 1

The sound in a loud part of the room is 10000 times louder than sound in a quiet part of the same place.

Explanation:

The acoustic intensity sound is a logarithmic function whose form is:

`L = 10\cdot \log_(10)\left((I)/(I_(o)) \right)`

Where:

`L` - Acoustic intensity sound, measured in decibels.

`I_(o)` - Reference sound intensity, measured in watts per square meter.

`I` - Real sound intensity, measured in watts per square meter.

Sound intensity is now cleared:

`10^{(L)/(10) } = (I)/(I_(o))`

The ratio of the sound intensity in a loud part to the sound intensity in a quiet part is:

`(I_(100))/(I_(60)) = \frac{10^{(100\,dB)/(10) }}{10^{(60\,dB)/(10)}}`

`(I_(100))/(I_(60)) = \left(10^(100\,dB-60\,dB)\right)^{(1)/(10) }`

`(I_(100))/(I_(60)) = (10^(40\,dB))^{(1)/(10) }`

`(I_(100))/(I_(60)) =10^(4)`

The sound in a loud part of the room is 10000 times louder than sound in a quiet part of the same place.