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Chapter 16: Problem 53

A loudspeaker has a circular opening with a radius of \(0.0950 \mathrm{~m} .\) The electrical power needed to operate the speaker is \(25.0 \mathrm{~W}\). The average sound intensity at the opening is \(17.5 \mathrm{~W} / \mathrm{m}^{2}\). What percentage of the electrical power is converted by the speaker into sound power?

Short Answer

Expert verified
About 1.981% of the electrical power is converted into sound power.

Step by step solution

01

Calculate the Area of the Speaker's Opening

The first step is to calculate the area of the circular opening of the speaker using the formula for the area of a circle, which is \( A = \pi r^2 \). Given the radius \( r = 0.0950 \; \text{m} \), we substitute the value into the formula:\[ A = \pi (0.0950)^2 \approx 0.0283 \; \text{m}^2 \]
02

Determine the Sound Power Produced by the Speaker

Using the sound intensity \( I = 17.5 \; \text{W/m}^2 \) and the area of the opening \( A \), the sound power \( P_{\text{sound}} \) can be calculated by multiplying the intensity by the area:\[ P_{\text{sound}} = I \times A = 17.5 \; \text{W/m}^2 \times 0.0283 \; \text{m}^2 \approx 0.49525 \; \text{W} \]
03

Calculate the Percentage of Electrical Power Converted to Sound

The percentage of electrical power converted into sound power is given by dividing the sound power \( P_{\text{sound}} \) by the total electrical power \( P_{\text{electrical}} = 25.0 \; \text{W} \) and then multiplying by 100%:\[ \text{Percentage} = \left( \frac{P_{\text{sound}}}{P_{\text{electrical}}} \right) \times 100 = \left( \frac{0.49525}{25.0} \right) \times 100 \approx 1.981 \% \]

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Key Concepts

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

Circular Opening Area
To understand how a loudspeaker works, we first need to know the area of its circular opening. Calculating this area helps us determine how much sound the speaker can project through its surface. The formula for the area of a circle is simple:
  • \( A = \pi r^2 \)
Here, \( A \) is the area, \( \pi \) is a constant approximately equal to 3.14159, and \( r \) is the radius of the circle.
In the case of the loudspeaker from our exercise, the radius is given as \( r = 0.0950 \; \text{m} \). By plugging this value into the formula, we get:
  • \( A = \pi \times (0.0950)^2 \approx 0.0283 \; \text{m}^2 \)
This tells us the effective area through which sound waves are emitted from the speaker into the surrounding air.
Sound Power Calculation
Once the area of the loudspeaker's opening is known, the next step is to find out how much sound power it emits. Sound power is different from sound intensity, although they are related.
Sound intensity is defined as the sound power per unit area, and it's given in this exercise as \( 17.5 \; \text{W/m}^2 \). To determine the total sound power \( P_{\text{sound}} \), we multiply the intensity by the area:
  • \( P_{\text{sound}} = I \times A = 17.5 \; \text{W/m}^2 \times 0.0283 \; \text{m}^2 \approx 0.49525 \; \text{W} \)
This calculation shows us how much of the power from the speaker is actually being converted into sound that can be heard鈥攁 crucial factor in assessing speaker performance.
Electrical to Sound Power Conversion
The final step involves understanding the efficiency of the loudspeaker, which is how well it converts electrical power to sound power. The speaker uses an electrical power input of \( 25.0 \; \text{W} \), but not all of this power is transformed into sound.
To find the conversion efficiency, or what percentage of the electrical power is converted, we use the formula:
  • \( \text{Percentage} = \left( \frac{P_{\text{sound}}}{P_{\text{electrical}}} \right) \times 100 \)
Substituting the values from the exercise, we calculate:
  • \( \text{Percentage} = \left( \frac{0.49525}{25.0} \right) \times 100 \approx 1.981 \% \)
This means that only about 1.981 percent of the electric power is converted into sound. The rest may be lost as heat or other forms of energy. This conversion percentage helps evaluate the speaker's efficiency.

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Most popular questions from this chapter

A person lying on an air mattress in the ocean rises and falls through one complete cycle every five seconds. The crests of the wave causing the motion are \(20.0 \mathrm{~m}\) apart. Determine (a) the frequency and (b) the speed of the wave.

A portable radio is sitting at the edge of a balcony \(5.1 \mathrm{~m}\) above the ground. The unit is emitting sound uniformly in all directions. By accident, it falls from rest off the balcony and continues to play on the way down. A gardener is working in a flower bed directly below the falling unit. From the instant the unit begins to fall, how much time is required for the sound intensity level heard by the gardener to increase by \(10.0 \mathrm{~dB} ?\)

Review Interactive Solution \(\underline{16.55}\) at for one approach to this problem. A dish of lasagna is being heated in a microwave oven. The effective area of the lasagna that is exposed to the microwaves is \(2.2 \times 10^{-2} \mathrm{~m}^{2}\). The mass of the lasagna is \(0.35 \mathrm{~kg},\) and its specific heat capacity is \(3200 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). The temperature rises by \(72 \mathrm{C}^{\circ}\) in 8.0 minutes. What is the intensity of the microwaves in the oven?

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{~kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C}\), how many times greater is the speed of sound in the rail than in the air?

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of \(240 \mathrm{~m} / \mathrm{s}\). What would be the speed of the wave on the longer wire?
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