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Chapter 12: Problem 17

(a) Show that if \(I_{2}=10.01_{1},\) then $\beta_{2}=\beta_{1}+10.0 \mathrm{dB} .$ (A factor of 10 increase in intensity corresponds to a 10.0 dB increase in intensity level.) (b) Show that if \(I_{2}=2.0 I_{1}\) then \(\beta_{2}=\beta_{1}+3.0 \mathrm{dB} .\) (A factor of 2 increase in intensity corresponds to a 3.0 -dB increase in intensity level. tutorial: decibels)

Short Answer

Expert verified
Question: Show the following relationships between intensity and intensity level in decibels: a) If \(I_{2} = 10 I_{1}\), then \(\beta_{2} = \beta_{1} + 10.0 dB\). b) If \(I_2 = 2 I_1\) then \(\beta_2 = \beta_1 + 3.0 dB\). Answer: a) We have proved that if \(I_{2} = 10 I_{1}\), then \(\beta_{2} = \beta_{1} + 10.0\,\mathrm{dB}\). b) We have proved that if \(I_2 = 2 I_1\), then \(\beta_2 = \beta_1 + 3.0\,\mathrm{dB}\).

Step by step solution

01

Write the intensity level formula for \(I_1\) and \(I_2\)

First, we write the decibel intensity formula for \(I_1\) and \(I_2\): \(\beta_1 = 10\log_{10}\left(\frac{I_1}{I_0}\right)\) \(\beta_2 = 10\log_{10}\left(\frac{I_2}{I_0}\right)\)
02

Substitute the given value \(I_2 = 10 I_1\)

Now, let's substitute the value \(I_2 = 10 I_1\) into the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(\frac{10 I_1}{I_0}\right)\)
03

Simplify the expression inside the logarithm

We can simplify the expression inside the logarithm in the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(10\frac{I_1}{I_0}\right)\)
04

Use the logarithm rules

According to logarithm rules, for any positive numbers x and any positive integer n, \(\log (x^n) = n\log(x)\). So we can simplify the expression further: \(\beta_2 = 10\log_{10}(10) + 10\log_{10}\left(\frac{I_1}{I_0}\right)\)
05

Calculate the logarithm and compare with \(\beta_1\)

Since \(\log_{10}(10) = 1\), we can find the value of \(\beta_2\) and compare it with \(\beta_1\): \(\beta_2 = 10 \cdot 1 + 10\log_{10}\left(\frac{I_1}{I_0}\right)\) So we have proved that: \(\beta_2 = \beta_1 + 10.0\,\mathrm{dB}\) #b) Show that if \(I_2 = 2 I_1\) then \(\beta_2 = \beta_1 + 3.0 dB\)#
06

Write the intensity level formula for \(I_1\) and \(I_2\)

First, we write the decibel intensity formula for \(I_1\) and \(I_2\): \(\beta_1 = 10\log_{10}\left(\frac{I_1}{I_0}\right)\) \(\beta_2 = 10\log_{10}\left(\frac{I_2}{I_0}\right)\)
07

Substitute the given value \(I_2 = 2 I_1\)

Now, let's substitute the value \(I_2 = 2 I_1\) into the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(\frac{2 I_1}{I_0}\right)\)
08

Simplify the expression inside the logarithm

We can simplify the expression inside the logarithm in the formula for \(\beta_2\): \(\beta_2 = 10\log_{10}\left(2\frac{I_1}{I_0}\right)\)
09

Use the logarithm rules

According to logarithm rules, for any positive numbers x and z, \(\log (x z) = \log(x) + \log(z)\). So we can simplify the expression further: \(\beta_2 = 10\log_{10}(2) + 10\log_{10}\left(\frac{I_1}{I_0}\right)\)
10

Calculate the logarithm and compare with \(\beta_1\)

Since \(\log_{10}(2) \approx 0.3010\), we can find the value of \(\beta_2\) and compare it with \(\beta_1\): \(\beta_2 = 10 \cdot 0.3010 + 10\log_{10}\left(\frac{I_1}{I_0}\right)\) So we have proved that: \(\beta_2 = \beta_1 + 3.0\,\mathrm{dB}\)

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