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Chapter 11: Problem 3

The intensity of the sound wave from a jet airplane as it is taking off is \(1.0 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) at a distance of $5.0 \mathrm{m}$ What is the intensity of the sound wave that reaches the ears of a person standing at a distance of \(120 \mathrm{m}\) from the runway? Assume that the sound wave radiates from the airplane equally in all directions.

Short Answer

Expert verified
Answer: The intensity of the sound wave that reaches the person standing 120 meters away from the runway is approximately \(0.1736 \mathrm{W/m^2}\).

Step by step solution

01

Inverse square law formula

We will use the inverse square law formula to relate the initial and final intensities: \(\frac{I1}{I2}\) = \(\frac{d2^2}{d1^2}\)
02

Solve for final intensity (I2)

Now, we will solve the equation for the final intensity (I2): \(I2\) = \(\frac{I1 脳 d1^2}{d2^2}\)
03

Substitute the given values into the equation

Plug the given values into the equation: \(I2\) = \(\frac{(1.0 脳 10^{2} \mathrm{W/m^2}) 脳 (5.0 \mathrm{m})^2}{(120 \mathrm{m})^2}\)
04

Calculate the answer

Now, calculate the intensity of the sound wave at 120 meters: \(I2\) = \(\frac{(1.0 脳 10^{2} \mathrm{W/m^2}) 脳 25 \mathrm{m^2}}{14400 \mathrm{m^2}}\) \(I2\) 鈮� \(\frac{2500 \mathrm{W/m^2}}{14400 \mathrm{m^2}}\) \(I2\) 鈮� \(0.1736 \mathrm{W/m^2}\)
05

Final answer

The intensity of the sound wave that reaches the ears of a person standing at a distance of \(120 \mathrm{m}\) from the runway is approximately \(0.1736 \mathrm{W/m^2}\).

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