Find the intensity in decibels $[I(\text{dB})]$ for each value of $I$.



$\begin{array}{ll}

\text{Normal conversation:} & I = 10^6 I_0 \\

& I(\text{dB}) = \\

\text{Power saw at } 3 \text{ feet:} & I = 10^{11} I_0 \\

& I(\text{dB}) = \\

\text{Jet engine at } 100 \text{ feet:} & I = 10^{18} I_0 \\

& I(\text{dB}) =

\end{array}$

Verified Answer :

  • The intensity level in decibels is calculated using the formula \(I(dB) = 10 \cdot \log_{10}(\frac{I}{I_0})\).
  • Using the logarithmic property \(\log_{10}(10^k) = k\), we simplify the ratio \(\frac{I}{I_0}\) to \(10^k\).
  • For normal conversation (\(I = 10^6 I_0\)), the result is \(10 \times 6 = \boxed{60 \text{ dB}}\).
  • For a power saw (\(I = 10^{11} I_0\)), the result is \(10 \times 11 = \boxed{110 \text{ dB}}\).
  • For a jet engine (\(I = 10^{18} I_0\)), the result is \(10 \times 18 = \boxed{180 \text{ dB}}\).

Explanation

  1. Analysis of the Problem and Formula

The intensity of a sound, in decibels (dB), is measured relative to a reference intensity \(I_0\). The formula to calculate the sound intensity level \(I(dB)\) is given by:
\[I(dB) = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)\]
where \(I\) is the intensity of the sound and \(I_0\) is the reference intensity. We use the logarithmic property \(\log_{10}(10^k) = k\) to simplify the calculations for each given scenario.

  1. Normal Conversation Intensity

For normal conversation, the intensity is given as \(I = 10^6 I_0\). Substituting this into the formula, we have:
\[I(dB) = 10 \cdot \log_{10}\left(\frac{10^6 I_0}{I_0}\right) = 10 \cdot \log_{10}(10^6)\]
Applying the property \(\log_{10}(10^k) = k\), where \(k = 6\):
\[I(dB) = 10 \cdot 6 = 60 \text{ dB}\]

  1. Power Saw Intensity

For a power saw at 3 feet, the intensity is given as \(I = 10^{11} I_0\). Substituting this into the formula, we have:
\[I(dB) = 10 \cdot \log_{10}\left(\frac{10^{11} I_0}{I_0}\right) = 10 \cdot \log_{10}(10^{11})\]
Applying the property \(\log_{10}(10^k) = k\), where \(k = 11\):
\[I(dB) = 10 \cdot 11 = 110 \text{ dB}\]

  1. Jet Engine Intensity

For a jet engine at 100 feet, the intensity is given as \(I = 10^{18} I_0\). Substituting this into the formula, we have:
\[I(dB) = 10 \cdot \log_{10}\left(\frac{10^{18} I_0}{I_0}\right) = 10 \cdot \log_{10}(10^{18})\]
Applying the property \(\log_{10}(10^k) = k\), where \(k = 18\):
\[I(dB) = 10 \cdot 18 = 180 \text{ dB}\]

Examples

Sound intensity levels are crucial in environmental planning and occupational health. For example, knowing the decibel levels of construction equipment helps safety engineers determine the duration a worker can safely be exposed to noise without using hearing protection. By converting physical intensity measurements into decibels, we make it easier to understand and communicate the potential impact of different sounds on human hearing in everyday environments.