(a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What i

step 1 Hey everyone, this is question number 18 from Chapter 17. In this problem, we're given a 90 deci

(a) What is the decibel level of a sound that is twice as...

Key Concepts

Logarithmic Function in Acoustics

Logarithmic functions are used in acoustics to convert multiplicative changes in sound intensity into additive changes in decibel level. This approach allows for easy calculation of changes in sound levels when the intensity is increased or decreased by a factor, using the formula change in dB = 10 log10(factor).

Sound Intensity Ratio

Sound intensity ratio involves comparing the intensity of two sounds by calculating the ratio of their intensities. This concept is crucial when determining how changes in sound power affect the decibel level, as a doubling, halving, or any multiplicative change in intensity can be directly translated into a decibel increase or decrease using logarithms.

Decibel Scale

The decibel scale is a logarithmic unit used to express the ratio of a physical quantity, usually power or intensity. In acoustics, it is used to quantify sound intensity relative to a reference level. Because the ear responds logarithmically to changes in intensity, the decibel scale provides a more natural way to represent how loud a sound is perceived.

Transcript

00:01 Hey everyone, this is question number 18 from chapter 17.

00:04 In this problem, we're given a 90 decibel sound.

00:06 We have two parts.

00:07 Part a asked to find the decibel level of a sound that is twice as intense as this, and we're asked to find, in part, b, the decibel level of a sound that is one -fifth as intense as this.

00:19 So i went ahead and wrote at our intensity equation with decibels b equals 10 log i over i -0.

00:24 So let's go ahead and solve for the intensity of this decibel sound, this 90 -decible sound.

00:29 So we're going to go, i'm just going to skip the step, divide 10 over the other side.

00:33 So we have 90 decibels over 10 equals log.

00:42 I over i not, which is our standard, 10 to the minus 12 watts per meter squared.

00:53 Okay, 90 divided by 10 is 9.

00:56 Then we're going to take the anti -log.

00:57 So raise 9 to the, raise 10 to the 9th power.

01:02 And then if we raise 10 to the log power, then that cancels the 10.

01:07 The log.

01:08 So now we have 10 to the 9th equals i over watts per meter squared.

01:29 Now we just multiply over and we can solve for intensity.

01:35 So that is going to give us an intensity of 1 times 10 to the minus 3 watts per meter squared.

01:51 All right, so that's our intensity for the 90 decibel sound.

01:54 So now part a, find decibel level of a sound that is twice as intense...

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