Example 8: Sound intensity I, measured in watts per square meter (W/m?), at a point r meters from a sound source with acoustic power P is given by 1(r) =* 4mr?
a) Find the sound intensity at two points r1 = 10m and r2 = 15m from a sound source with power P = 100W. Then find
the slope of the secant line through the points (10, 1(10)) and (15, 1(15)) and interpret the result.
b) Find the slope of the secant line through any two points (11, (r1)) and (12,1(r2)) on the graph of the intensity function with acoustic power P.

We are given the intensity function for sound intensity \( I(r) \) as a function of the distance \( r \) from a sound source, which is given by:

\[
I(r) = \frac{P}{4 \pi r^2}
\]

Where:
- \( I(r) \) is the sound intensity (in watts per square meter, W/m²),
- \( P \) is the acoustic power (in watts, W),
- \( r \) is the distance from the sound source (in meters),
- \( \pi \) is approximately 3.1416.

We will solve each part step by step.

### Part a: Sound Intensity at \( r_1 = 10 \, \text{m} \) and \( r_2 = 15 \, \text{m} \)

Given \( P = 100 \, \text{W} \), the intensity function becomes:

\[
I(r) = \frac{100}{4 \pi r^2}
\]

1. **At \( r_1 = 10 \, \text{m} \):**

\[
I(10) = \frac{100}{4 \pi (10^2)} = \frac{100}{4 \pi \cdot 100} = \frac{100}{400 \pi} = \frac{1}{4 \pi} \, \text{W/m}^2
\]
Approximating \( \pi \approx 3.1416 \):

\[
I(10) \approx \frac{1}{4 \times 3.1416} \approx \frac{1}{12.5664} \approx 0.0796 \, \text{W/m}^2
\]

2. **At \( r_2 = 15 \, \text{m} \):**

\[
I(15) = \frac{100}{4 \pi (15^2)} = \frac{100}{4 \pi \cdot 225} = \frac{100}{900 \pi} = \frac{1}{9 \pi} \, \text{W/m}^2
\]
Approximating \( \pi \):

\[
I(15) \approx \frac{1}{9 \times 3.1416} \approx \frac{1}{28.2744} \approx 0.0354 \, \text{W/m}^2
\]

### Slope of the Secant Line

The slope \( m \) of the secant line through the points \( (r_1, I(r_1)) \) and \( (r_2, I(r_2)) \) is given by the formula:

\[
m = \frac{I(r_2) - I(r_1)}{r_2 - r_1}
\]

Substitute the values for \( I(10) \), \( I(15) \), \( r_1 = 10 \), and \( r_2 = 15 \):

\[
m = \frac{0.0354 - 0.0796}{15 - 10} = \frac{-0.0442}{5} = -0.00884 \, \text{W/m}^2 \text{ per meter}
\]

### Interpretation of the Slope:

The negative slope indicates that the intensity decreases as the distance from the source increases, which is consistent with the inverse-square law of sound intensity. The rate of decrease in intensity is approximately \( 0.00884 \, \text{W/m}^2 \) per meter for the given range.

### Part b: General Slope of the Secant Line for Any \( r_1 \) and \( r_2 \)

The formula for the sound intensity is:

\[
I(r) = \frac{100}{4 \pi r^2}
\]

For two arbitrary points \( r_1 \) and \( r_2 \), the slope of the secant line is:

\[
m = \frac{I(r_2) - I(r_1)}{r_2 - r_1}
\]

Substituting the intensity function:

\[
m = \frac{\frac{100}{4 \pi r_2^2} - \frac{100}{4 \pi r_1^2}}{r_2 - r_1}
\]

Factor out \( \frac{100}{4 \pi} \):

\[
m = \frac{100}{4 \pi} \cdot \frac{\frac{1}{r_2^2} - \frac{1}{r_1^2}}{r_2 - r_1}
\]

This is the general formula for the slope of the secant line between any two points on the graph of the sound intensity function.

---

Would you like further details on any step? Here are 5 related questions for more exploration:

1. What is the physical meaning of the inverse-square law in sound propagation?
2. How does the secant line slope change for different distances \( r_1 \) and \( r_2 \)?
3. How is sound intensity affected by changes in acoustic power \( P \)?
4. How would the sound intensity change if the medium affected the propagation?
5. Can the slope of the secant line be used to estimate the sound intensity decay rate?

**Tip**: The inverse-square law is a principle that applies to many physical phenomena, including light and gravitational forces, not just sound.

Sound intensity I, measured in watts per square meter (W/m²), at a point r meters from a sound source with acoustic power P is given by I(r) = P / (4πr²). a) Find the sound intensity at two points r₁ = 10m and r₂ = 15m from a sound source with power P = 100W. Then find the slope of the secant line through the points (10, I(10)) and (15, I(15)) and interpret the result. b) Find the slope of the secant line through any two points (r₁, I(r₁)) and (r₂, I(r₂)) on the graph of the intensity function with acoustic power P.

Learn how to calculate sound intensity at distances of 10m and 15m from a 100W sound source using the inverse-square law. Find the slope of the secant line between these points and understand the rate of intensity decrease over distance. Suitable for advanced high school physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation