Answer :
Sure, let's solve each part step-by-step.
### Part (a)
We need to find the decibel rating of a sound with intensity [tex]\( I = 100,000 I_0 \)[/tex], where [tex]\( I_0 \)[/tex] is the initial intensity. The formula for decibel rating [tex]\( d \)[/tex] is given by:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Substituting the given intensity [tex]\( I = 100,000 I_0 \)[/tex] into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000) \][/tex]
We know that [tex]\( 100,000 \)[/tex] can be written as [tex]\( 10^5 \)[/tex], so:
[tex]\[ d = 10 \log (10^5) \][/tex]
Since [tex]\( \log (10^5) = 5 \)[/tex]:
[tex]\[ d = 10 \times 5 \][/tex]
[tex]\[ d = 50 \text{ decibels} \][/tex]
So, the decibel rating of the sound is [tex]\( 50 \)[/tex] decibels.
### Part (b)
Now, we need to determine the increase in decibel rating when the intensity of the sound is tripled. The new intensity [tex]\( I_{\text{new}} \)[/tex] is:
[tex]\[ I_{\text{new}} = 3 \times I \][/tex]
[tex]\[ I_{\text{new}} = 3 \times 100,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 300,000 I_0 \][/tex]
Using the same formula for decibel rating:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{300,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (300,000) \][/tex]
We know that [tex]\( 300,000 \)[/tex] can be written as [tex]\( 3 \times 10^5 \)[/tex], so:
[tex]\[ d_{\text{new}} = 10 \log (3 \times 10^5) \][/tex]
Using the properties of logarithms, we get:
[tex]\[ d_{\text{new}} = 10 \left( \log 3 + \log 10^5 \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \left( \log 3 + 5 \right) \][/tex]
Given that [tex]\( \log 3 \approx 0.4771 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \left( 0.4771 + 5 \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 5.4771 \][/tex]
[tex]\[ d_{\text{new}} \approx 54.77 \text{ decibels} \][/tex]
The increase in decibel rating is then:
[tex]\[ d_{\text{increase}} = d_{\text{new}} - d \][/tex]
[tex]\[ d_{\text{increase}} = 54.77 - 50 \][/tex]
[tex]\[ d_{\text{increase}} \approx 4.77 \text{ decibels} \][/tex]
Rounding to the nearest integer, the increase in decibel rating is approximately [tex]\( 5 \)[/tex] decibels.
### Summary:
(a) The decibel rating of the sound having intensity [tex]\( 100,000 I_0 \)[/tex] is [tex]\( 50 \)[/tex] decibels.
(b) If the intensity of the sound is tripled, the decibel rating increases by about [tex]\( 5 \)[/tex] decibels.
### Part (a)
We need to find the decibel rating of a sound with intensity [tex]\( I = 100,000 I_0 \)[/tex], where [tex]\( I_0 \)[/tex] is the initial intensity. The formula for decibel rating [tex]\( d \)[/tex] is given by:
[tex]\[ d = 10 \log \left( \frac{I}{I_0} \right) \][/tex]
Substituting the given intensity [tex]\( I = 100,000 I_0 \)[/tex] into the formula:
[tex]\[ d = 10 \log \left( \frac{100,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d = 10 \log (100,000) \][/tex]
We know that [tex]\( 100,000 \)[/tex] can be written as [tex]\( 10^5 \)[/tex], so:
[tex]\[ d = 10 \log (10^5) \][/tex]
Since [tex]\( \log (10^5) = 5 \)[/tex]:
[tex]\[ d = 10 \times 5 \][/tex]
[tex]\[ d = 50 \text{ decibels} \][/tex]
So, the decibel rating of the sound is [tex]\( 50 \)[/tex] decibels.
### Part (b)
Now, we need to determine the increase in decibel rating when the intensity of the sound is tripled. The new intensity [tex]\( I_{\text{new}} \)[/tex] is:
[tex]\[ I_{\text{new}} = 3 \times I \][/tex]
[tex]\[ I_{\text{new}} = 3 \times 100,000 I_0 \][/tex]
[tex]\[ I_{\text{new}} = 300,000 I_0 \][/tex]
Using the same formula for decibel rating:
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{I_{\text{new}}}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log \left( \frac{300,000 I_0}{I_0} \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \log (300,000) \][/tex]
We know that [tex]\( 300,000 \)[/tex] can be written as [tex]\( 3 \times 10^5 \)[/tex], so:
[tex]\[ d_{\text{new}} = 10 \log (3 \times 10^5) \][/tex]
Using the properties of logarithms, we get:
[tex]\[ d_{\text{new}} = 10 \left( \log 3 + \log 10^5 \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \left( \log 3 + 5 \right) \][/tex]
Given that [tex]\( \log 3 \approx 0.4771 \)[/tex]:
[tex]\[ d_{\text{new}} = 10 \left( 0.4771 + 5 \right) \][/tex]
[tex]\[ d_{\text{new}} = 10 \times 5.4771 \][/tex]
[tex]\[ d_{\text{new}} \approx 54.77 \text{ decibels} \][/tex]
The increase in decibel rating is then:
[tex]\[ d_{\text{increase}} = d_{\text{new}} - d \][/tex]
[tex]\[ d_{\text{increase}} = 54.77 - 50 \][/tex]
[tex]\[ d_{\text{increase}} \approx 4.77 \text{ decibels} \][/tex]
Rounding to the nearest integer, the increase in decibel rating is approximately [tex]\( 5 \)[/tex] decibels.
### Summary:
(a) The decibel rating of the sound having intensity [tex]\( 100,000 I_0 \)[/tex] is [tex]\( 50 \)[/tex] decibels.
(b) If the intensity of the sound is tripled, the decibel rating increases by about [tex]\( 5 \)[/tex] decibels.
