Question

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I isβ=10log(II0)dB,where I0 is a reference intensity. For sound waves, I0 is taken to be 10−12W/m2. Note that log refers to the logarithm to the base 10.Part AWhat is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity (i.e., I=10I0)?Part BWhat is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity (i.e. I=100I0)?Express the sound intensity numerically to the nearest integer.

Answer 1

Most of the information to solve this problem is provided in the statement, therefore we will apply the concepts related to the intensity of the sound and its method of representation across the logarithmic scale.

By definition as we saw the level of sound intensity in decibels is represented by

`\beta = 10log((I)/(I_0))dB`

Where, I = Intensity for which decibels is to be calculated

`I_0`= Reference intensity (at this case is
`10^(-12)W/m^2`

PART A ) Intensity is 10 times the reference intensity.

Here
`I = 10I_0`, replacing

`\beta = 10log((10I_0)/(I_0))dB`

`\beta = 10log(10)dB`

`\beta = 10dB`

Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 10dB

PART B) Intensity is 100 times the reference intensity

Here
`I = 100I_0`, replacing

`\beta = 10log((100I_0)/(I_0))dB`

`\beta = 10log(100)dB`

`\beta = 20dB`

Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 20dB