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Sagot :
Sure, let's break down the solution to the problem step by step.
### Part (a): Finding the Decibel Rating
To find the decibel rating [tex]\( d \)[/tex] of a sound with an intensity [tex]\( 100,000,000,000 I_0 \)[/tex] (where [tex]\( I_0 \)[/tex] is the initial intensity), we use the formula:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Here:
- [tex]\( I \)[/tex] is the given intensity.
- [tex]\( I_0 \)[/tex] is the initial intensity, which we assume to be normalized to 1 for simplicity.
Given:
[tex]\[ I = 100,000,000,000 I_0 \][/tex]
[tex]\[ I_0 = 1 \][/tex]
Substitute the values into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000,000,000) \][/tex]
Since [tex]\( \log (100,000,000,000) \)[/tex] can be computed as the log of [tex]\( 10^{11} \)[/tex]:
[tex]\[ d = 10 \log (10^{11}) \][/tex]
Using the logarithmic property that [tex]\( \log (10^x) = x \)[/tex]:
[tex]\[ d = 10 \times 11 \][/tex]
[tex]\[ d = 110 \][/tex]
Thus, the decibel rating [tex]\( d \)[/tex] is:
[tex]\[ \boxed{110.0 \text{ dB}} \][/tex]
### Part (b): Increase in Decibel Rating When Intensity is Doubled
If the intensity of the sound is doubled, the new intensity [tex]\( I_{\text{new}} \)[/tex] becomes:
[tex]\[ I_{\text{new}} = 2 \times 100,000,000,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 200,000,000,000 I_0 \][/tex]
We need to find the new decibel rating [tex]\( d_{\text{new}} \)[/tex] using the same formula:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
Substitute the new intensity [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{200,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (200,000,000,000) \][/tex]
Recognize that [tex]\( 200,000,000,000 \)[/tex] can be written as [tex]\( 2 \times 10^{11} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log (2 \times 10^{11}) \][/tex]
Using the property of logarithms [tex]\( \log (ab) = \log (a) + \log (b) \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + \log (10^{11})) \][/tex]
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Knowing [tex]\( \log 10^{11} = 11 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Given that [tex]\( \log 2 \approx 0.3010 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (0.3010 + 11) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 11.3010 \][/tex]
[tex]\[ d_{\text{new}} = 113.010 \][/tex]
Thus, the new decibel rating [tex]\( d_{\text{new}} \)[/tex] is:
[tex]\[ \boxed{113.0103 \text{ dB}} \][/tex]
Finally, the increase in the decibel rating when the intensity is doubled is:
[tex]\[ \Delta d = d_{\text{new}} - d \][/tex]
[tex]\[ \Delta d = 113.0103 \text{ dB} - 110.0 \text{ dB} \][/tex]
[tex]\[ \Delta d = 3.0103 \text{ dB} \][/tex]
So, the increase in the decibel rating is:
[tex]\[ \boxed{3.0103 \text{ dB}} \][/tex]
To summarize:
- The decibel rating for an intensity of [tex]\( 100,000,000,000 I_0 \)[/tex] is [tex]\( 110.0 \text{ dB} \)[/tex].
- When the intensity is doubled, the decibel rating increases by [tex]\( 3.0103 \text{ dB} \)[/tex].
### Part (a): Finding the Decibel Rating
To find the decibel rating [tex]\( d \)[/tex] of a sound with an intensity [tex]\( 100,000,000,000 I_0 \)[/tex] (where [tex]\( I_0 \)[/tex] is the initial intensity), we use the formula:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Here:
- [tex]\( I \)[/tex] is the given intensity.
- [tex]\( I_0 \)[/tex] is the initial intensity, which we assume to be normalized to 1 for simplicity.
Given:
[tex]\[ I = 100,000,000,000 I_0 \][/tex]
[tex]\[ I_0 = 1 \][/tex]
Substitute the values into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000,000,000) \][/tex]
Since [tex]\( \log (100,000,000,000) \)[/tex] can be computed as the log of [tex]\( 10^{11} \)[/tex]:
[tex]\[ d = 10 \log (10^{11}) \][/tex]
Using the logarithmic property that [tex]\( \log (10^x) = x \)[/tex]:
[tex]\[ d = 10 \times 11 \][/tex]
[tex]\[ d = 110 \][/tex]
Thus, the decibel rating [tex]\( d \)[/tex] is:
[tex]\[ \boxed{110.0 \text{ dB}} \][/tex]
### Part (b): Increase in Decibel Rating When Intensity is Doubled
If the intensity of the sound is doubled, the new intensity [tex]\( I_{\text{new}} \)[/tex] becomes:
[tex]\[ I_{\text{new}} = 2 \times 100,000,000,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 200,000,000,000 I_0 \][/tex]
We need to find the new decibel rating [tex]\( d_{\text{new}} \)[/tex] using the same formula:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
Substitute the new intensity [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{200,000,000,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (200,000,000,000) \][/tex]
Recognize that [tex]\( 200,000,000,000 \)[/tex] can be written as [tex]\( 2 \times 10^{11} \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \log (2 \times 10^{11}) \][/tex]
Using the property of logarithms [tex]\( \log (ab) = \log (a) + \log (b) \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + \log (10^{11})) \][/tex]
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Knowing [tex]\( \log 10^{11} = 11 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (\log 2 + 11) \][/tex]
Given that [tex]\( \log 2 \approx 0.3010 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 (0.3010 + 11) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 11.3010 \][/tex]
[tex]\[ d_{\text{new}} = 113.010 \][/tex]
Thus, the new decibel rating [tex]\( d_{\text{new}} \)[/tex] is:
[tex]\[ \boxed{113.0103 \text{ dB}} \][/tex]
Finally, the increase in the decibel rating when the intensity is doubled is:
[tex]\[ \Delta d = d_{\text{new}} - d \][/tex]
[tex]\[ \Delta d = 113.0103 \text{ dB} - 110.0 \text{ dB} \][/tex]
[tex]\[ \Delta d = 3.0103 \text{ dB} \][/tex]
So, the increase in the decibel rating is:
[tex]\[ \boxed{3.0103 \text{ dB}} \][/tex]
To summarize:
- The decibel rating for an intensity of [tex]\( 100,000,000,000 I_0 \)[/tex] is [tex]\( 110.0 \text{ dB} \)[/tex].
- When the intensity is doubled, the decibel rating increases by [tex]\( 3.0103 \text{ dB} \)[/tex].
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