Question

The sound level measured in a room by a person watching a movie on a home theater system varies from 50 dB during a quiet part to 70 dB during a loud part. Approximately how many times louder is the latter sound

Answer 1

A 70 dB sound is 100 times more intense than a 50 dB sound because a 10 dB increase represents a tenfold increase in intensity, and there is a 20 dB difference between the two sounds.

Step-by-step explanation:

• Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.
• SI unit of Sound Intensity is called Decibel (dB).

• The question concerns how many times louder a 70 dB sound is compared to a 50 dB sound.
• Since the decibel scale is a logarithmic measure of sound intensity, a difference of 20 dB means that the 70 dB sound is 100 times more intense than the 50 dB sound (10 dB difference corresponds to a factor of 10 in intensity).

Answer 2

100 times

Step-by-step explanation:

The sound intensity level β of a sound with an intensity I is mathematically given as:

`\beta (dB)=10log_(10)((I)/(I_0) )`, Where
`I_0` = lowest sound intensity for a normal person at a frequency of 1000 Hz

For the quiet part:

`\beta_1 = 10*log_(10)*(I_1/I_0)`

For the loud part:

`\beta_2 = 10*log_(10) * (I_2/I_0)`

Hence,

`\beta_2 - \beta_1 = 10*log_(10) * (I_2/I_1)`

70-50 =
`10*log_(10) * (I_2/I_1)`

`log_(10) * (I_2/I_1)` = 2

`(I_2/I_1)` = 100

`I_2` = 100
`I_1`

Therefore, the latter sound (
`I_2`) is 100 times louder than the former sound (
`I_1`)