Question

Workers around jet aircraft typically wear protective devices over their ears. Assume that the sound level of a jet airplane engine, at a distance of 29 m, is 130 dB, and that the average human ear has an effective radius of 1.9 cm. What would be the power intercepted by an unprotected ear at a distance of 29 m from a jet airplane engine?

Answer 1

The power intercepted by an unprotected ear, exposed to a jet airplane engine at 29 m, is approximately
`\(1.13 * 10^(-3)\)` watts, calculated from the sound intensity and ear area.

To calculate the power intercepted by an unprotected ear, we can use the formula for sound intensity and then relate it to power.

The formula for sound intensity is given by:

`\[ I = (P)/(A) \]`

where:

- (I) is the sound intensity,

- (P) is the power, and

- (A) is the area through which the sound is passing.

The area (A) can be calculated using the formula for the area of a circle:

`\[ A = \pi r^2 \]`

Given:

- Sound level (L) = 130 dB

- Distance (r) = 29 m

- Ear radius
`(\(r_{\text{ear}}\))` = 1.9 cm = 0.019 m (converted to meters)

First, we need to find the sound intensity
`(\(I\))` using the formula:

`\[ L = 10 \cdot \log_(10)\left((I)/(I_0)\right) \]`

where
`\(I_0\)` is the reference intensity (typically
`\(1.0 * 10^(-12) \, \text{W/m}^2\)` for sound).

1. Solve for I:

`\[ I = I_0 \cdot 10^((L/10)) \]`

2. Calculate the area (A) of the ear using
`\(\pi r_{\text{ear}}^2\)`.

3. Now, calculate the power (P) intercepted by the ear using
`\(I = (P)/(A)\)`.

Let's go through the calculations:

`\[ I = (1.0 * 10^(-12) \, \text{W/m}^2) \cdot 10^((130/10)) \]\[ I \approx 1.0 \, \text{W/m}^2 \]\[ A = \pi \cdot (0.019 \, \text{m})^2 \]\[ A \approx 1.13 * 10^(-3) \, \text{m}^2 \]\[ P = I \cdot A \]\[ P \approx (1.0 \, \text{W/m}^2) \cdot (1.13 * 10^(-3) \, \text{m}^2) \]\[ P \approx 1.13 * 10^(-3) \, \text{W} \]`

Therefore, the power intercepted by an unprotected ear at a distance of 29 m from a jet airplane engine is approximately
`\(1.13 * 10^(-3)\)` watts.