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The sound level, $\beta$, in decibels (dB), of a sound with an intensity of $I$ is calculated in relation to the threshold of human hearing, $I_0$, by this equation:

$\beta = 10 \log \left(\frac{I}{I_0}\right)$



The threshold of human hearing is $10^{-12}\text{ watts/meter}^2$. The sound level of a jet plane is approximately $140\text{ dB}$. The intensity of the sound of a jet plane is approximately [____] $\text{watts/meter}^2$.



Options:

  • 10,000
  • 10
  • 100
  • 1,000

Verified Answer :

  • The problem defines the sound intensity formula as \(\beta = 10 \log \left(\frac{I}{I_0}\right)\).
  • We substitute \(\beta = 140\) and \(I_0 = 10^{-12}\) into the equation.
  • Solving the equation leads to \(\log_{10} \left(\frac{I}{10^{-12}}\right) = 14\), which implies \(\frac{I}{10^{-12}} = 10^{14}\).
  • Multiplying gives \(I = 10^{14} \cdot 10^{-12} = 10^2 = \boxed{100}\).

Explanation

  1. Data and Problem Analysis

The problem provides the formula for sound intensity level in decibels (dB), which is \(\beta = 10 \log \left(\frac{I}{I_0}\right)\). Here, \(\beta\) represents the sound level, \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity corresponding to the threshold of human hearing. We are given:

  • The sound level of a jet plane: \(\beta = 140 \text{ dB}\)
  • The threshold of human hearing: \(I_0 = 10^{-12} \text{ watts/meter}^2\)

We need to find the intensity \(I\) of the jet plane's sound.

  1. Isolating and Simplifying the Logarithmic Equation

To find \(I\), we substitute the given values into the formula:
\[140 = 10 \log \left(\frac{I}{10^{-12}}\right)\]
First, we divide both sides by 10 to isolate the logarithmic term:
\[\frac{140}{10} = \log \left(\frac{I}{10^{-12}}\right)\]
\[14 = \log_{10} \left(\frac{I}{10^{-12}}\right)\]
Using the definition of a logarithm, which states that \(\log_{b}(x) = y\) is equivalent to \(b^y = x\), we can rewrite the equation as:
\[10^{14} = \frac{I}{10^{-12}}\]

  1. Solving for Intensity and Conclusion

Now, we solve for \(I\) by multiplying both sides of the equation by \(10^{-12}\):
\[I = 10^{14} \times 10^{-12}\]
Applying the laws of exponents, specifically \(a^m \times a^n = a^{m+n}\), we combine the exponents:
\[I = 10^{14 + (-12)}\]
\[I = 10^{2}\]
Calculating the value of \(10^2\), we get:
\[I = 100\]
Therefore, the intensity of the sound of a jet plane is \(\boxed{100}\) \(\text{watts/meter}^2\).

Examples

Sound intensity is essential for safety, especially in industrial or aviation settings. For example, airports use these calculations to determine noise exposure zones for workers and nearby residents. By understanding how decibels relate to actual power intensity, planners can design sound barriers, define safe distance zones for ground crew, and ensure compliance with occupational noise exposure regulations, ultimately protecting human hearing from long-term damage.

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