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Chapter 18: Problem 24

A point source emits \(30.0 \mathrm{~W}\) of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 200 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

Short Answer

Expert verified
Intensity at 200 m is 5.97 脳 10鈦烩伒 W/m虏. Power intercepted by the microphone is 4.48 脳 10鈦烩伖 W.

Step by step solution

01

Understanding the Problem

A point source emits sound power of 30.0 W in all directions. A microphone intercepts this sound over an area of 0.750 cm虏 at a distance of 200 meters from the source. We need to calculate (a) the sound intensity at 200 meters and (b) the power intercepted by the microphone.
02

Convert the Microphone Area to Square Meters

Convert the area from square centimeters to square meters by using the conversion factor:\[ 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \]Therefore,\[ 0.750 \text{ cm}^2 = 0.750 \times 10^{-4} \text{ m}^2 = 7.50 \times 10^{-5} \text{ m}^2 \]
03

Calculate the Surface Area of the Sphere

The sound spreads out uniformly in all directions, forming a sphere. The surface area of a sphere is given by:\[ A = 4 \pi r^2 \]where \( r \) is the radius (distance from the source). So with \( r = 200 \text{ m} \):\[ A = 4 \pi (200 \text{ m})^2 = 160,000 \pi \text{ m}^2 \]
04

Calculate the Intensity of Sound

Sound intensity (I) is the power (P) per unit area (A). Using the surface area calculated:\[ I = \frac{P}{A} \]Substitute the given power (P = 30.0 W) and the surface area:\[ I = \frac{30.0 \text{ W}}{160,000 \pi \text{ m}^2} = \frac{30.0}{160,000 \pi} \text{ W/m}^2 = 5.97 \times 10^{-5} \text{ W/m}^2 \]
05

Calculate the Power Intercepted by the Microphone

The power intercepted by the microphone (P_{intercepted}) is given by:\[ P_{intercepted} = I \times A_{microphone} \]Using the intensity calculated in Step 4 and the area of the microphone from Step 2:\[ P_{intercepted} = (5.97 \times 10^{-5} \text{ W/m}^2) \times (7.50 \times 10^{-5} \text{ m}^2) = 4.48 \times 10^{-9} \text{ W} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound Power
Sound power is the total amount of energy a sound source emits per second in all directions. It is measured in watts (W). In our problem, the point source emits 30.0 W of sound power uniformly. This type of source is called an isotropic emitter, meaning it disperses sound equally in all directions. Sound power is an important concept as it influences how intense the sound is at different distances.
Surface Area of a Sphere
When sound is emitted from a point source, it spreads out uniformly to form a spherical wavefront. The surface area of this sphere increases with the square of the radius (distance from the source), and is calculated using the formula: \[ A = 4 \pi r^2 \].Here, \( r \) is the distance from the source. Since the sound spreads out over this surface area, understanding this concept helps us determine how diluted the sound becomes as it moves away from the source. For instance, at a distance of 200 meters, the area is massive 鈥� about 160,000\( \pi \) square meters. As the radius grows, the same amount of power is distributed over a larger area, reducing the sound intensity proportionally.
Intensity of Sound
Sound intensity is the amount of sound power passing through a unit area perpendicular to the direction of sound propagation. It essentially measures the 'strength' of the sound at a specific point and is expressed in watts per square meter (W/m虏). The formula for calculating sound intensity is: \[ I = \frac{P}{A} \],where \( P \) is the power and \( A \) is the surface area over which the sound spreads. Given a sound power of 30.0 W and using the surface area calculated earlier, the intensity at 200 meters is very low, around 5.97 脳 10^{-5} W/m虏, because the sound power is spread over a large area.
Microphone Area Conversion
When dealing with sound interception by a microphone, the area of the microphone in square centimeters needs to be converted to square meters for consistency in units. This is done using the conversion factor: 1 cm虏 = 10^{-4} m虏. For example, the area of 0.750 cm虏 converts to 7.50 脳 10^{-5} m虏. This conversion is crucial for calculating the intercepted power accurately. The intercepted power is the product of sound intensity and the area of the microphone: \[ P_{intercepted} = I \times A_{microphone} \].Using the intensity calculated earlier and the converted area, we find that the power intercepted by the microphone is extremely small, about 4.48 脳 10^{-9} W.

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Most popular questions from this chapter

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