Solve each problem. Find the decibel ratings of sounds having the following intensities. (a) 100

Solve each problem. Find the decibel ratings of sounds...

Key Concepts

Sound Intensity Level and Decibels

This concept explains that the decibel scale is a logarithmic measure of the ratio of a given sound intensity to a reference intensity. The formula used is dB = 10 log10(I/I?), where I is the sound intensity and I? is the reference intensity. This scale is used because human hearing perceives sound logarithmically, allowing a wide range of intensities to be expressed in a more manageable form.

Logarithms

Logarithms are mathematical operations that convert multiplicative relationships into additive ones. In the context of sound intensity and decibels, logarithms simplify the process of scaling large differences in intensity into a smaller, interpretable scale by using the logarithm of the ratio of intensities.

Reference Intensity (I?)

The reference intensity (I?) is a standard value, often set at the threshold of human hearing (typically 10?¹² W/m²), against which other sound intensities are compared. It provides a baseline for calculating the decibel level, ensuring consistency in how sound levels are reported.

Doubling Sound Intensity

Doubling the sound intensity causes the decibel level to increase by a fixed amount due to the logarithmic nature of the scale. Mathematically, doubling the intensity results in an increase of about 10 log10(2) ? 3 decibels. This shows how multiplicative increases in intensity translate into additive increases in decibel ratings.

Transcript

00:01 Okay, so here, the decibel ratings of sounds is found by using this formula, where we have d is equal to 10 times log of i over i not.

00:12 Where here, i not is a threshold sound, and i is the intensity of a louder sound.

00:22 So then for part a, we want to find d when i is 100 times i not.

00:28 Please go ahead and substitute that into our formula and we have d equal to 100 times log of 100 i not over over i not okay well here i not is going to cancel out right i not over i not is just one so that cancels out and therefore just have d is equal to 110 times log of 100 100 or 100 or 10 times log of 100 or 100 or we know that log of 100 is just equal to log of 10 squared.

01:08 What is log of 10 squared? well, that's just 2, right? a log is an exponent.

01:16 So then, therefore, we substitute 2 in for log of 100 and simplify.

01:22 And then we have d equal to just 10 times 2, which is equal to 20.

01:28 So, therefore, the decimal rating of sound with intensity, 100 i .0 is equal to 20.

01:44 Then for b, we find d when i is 1 ,000 i0.

01:49 So now we substitute in 1 ,000 i not for i in our formula.

01:56 So then we have, so now this is part b.

02:00 Then we have d equal to 10 times log of 1 ,000 i .0 over i not.

02:12 Again, we have the i .naud cancel out.

02:15 So we simplify here.

02:16 We just get d is equal to 10 times log of 1 ,000.

02:20 That's 10 equal to log of 10 cubed, right? so we have d equal to 10 times log of 10 cubed.

02:31 10 cubed, well then the exponent here can come out front of the log as a coefficient.

02:36 So therefore, we have this equal to 10 times 3, log of 10.

02:43 Okay, but log of 10 is just 1.

02:46 So therefore, this just becomes 10 times 3, which is equal to 30.

02:53 So, therefore, the desperate rating of sound with intensity 1 ,000 i.

02:58 Knot is 30 and then for c we are going to substitute um 100 ,000 i not in 4 i so then we have d equal to 10 times log of well 100 ,000 i0 over i not again i not cancel out and we just get d equal to 10 times log of well 100 ,000 i knot over i not again i not cancel out and we just get d equal to 10 times log of a hundred thousand so that's the same thing as ten times log of ten to the fifth so therefore we have d is equal to ten times so that five comes out front so ten times five log of um log of ten is just one so therefore we just have this just equal thus this is just equal to ten times five which is 50...

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