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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
The speaker converts approximately 1.99% of electrical power into sound power.

Step by step solution

01

Calculate the area of the loudspeaker opening

To find the area of the circular opening, use the formula for the area of a circle: \( A = \pi r^2 \), where \( r = 0.0950 \, \text{m} \). Substitute the radius into the formula:\[ A = \pi (0.0950)^2 \approx 0.02836 \, \text{m}^2 \]
02

Calculate the sound power emitted by the speaker

The sound power \( P_s \) can be calculated using the formula for sound intensity \( I = \frac{P}{A} \), which can be rearranged to find \( P \) as follows: \( P = I \times A \). Substitute the known values \( I = 17.5 \, \text{W/m}^2 \) and the area from Step 1:\[ P_s = 17.5 \times 0.02836 \approx 0.4963 \text{ W} \]
03

Calculate the efficiency of the power conversion

Efficiency is calculated as the ratio of sound power to electrical power, expressed as a percentage. Use the formula: \( \text{Efficiency} = \left(\frac{P_s}{P_e}\right) \times 100 \% \). Substitute the sound power from Step 2 and the electrical power \( P_e = 25.0 \, \text{W} \):\[ \text{Efficiency} = \left(\frac{0.4963}{25.0}\right) \times 100 \% \approx 1.985 \% \]

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Key Concepts

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

Sound Intensity
Sound intensity is a measure of the sound power per unit area. It helps us understand how energy from a sound wave passes through a particular area. In more technical terms, sound intensity is the sound power, measured in watts (W), that reaches a surface, divided by the area of that surface, given in square meters (m²).

The formula to calculate sound intensity can be expressed as:
  • Sound Intensity (I) = \( \frac{P}{A} \)
where I is the intensity, P is the power in watts, and A is the area in square meters.

This concept is crucial in physics because it gives us an idea of how "loud" a sound is in terms of power distribution across an area, like the front of a loudspeaker. By knowing the sound intensity, we can also determine how efficiently a device, like a loudspeaker, converts electrical power into sound power. Understanding sound intensity is crucial in various applications, from designing concert halls to creating efficient loudspeakers.
Loudspeaker Physics
Loudspeakers are devices that convert electrical energy into sound. The physics behind loudspeakers involves several components working together to produce sound. At its core, a loudspeaker has a diaphragm, usually a cone-shaped object that moves to generate sound waves.

  • The diaphragm is attached to a coil of wire called the voice coil.
  • This coil surrounds a stationary magnet.
When an electrical current passes through the voice coil, it creates a magnetic field that interacts with the permanent magnet. This interaction causes the coil, and consequently the diaphragm, to move back and forth, producing sound waves in the air.

The shape and size of the loudspeaker affect the quality and intensity of the sound produced. A larger speaker can move more air, potentially producing louder sounds. On the other hand, the design and materials used also determine how well a loudspeaker can reproduce sound accurately. The radius of the loudspeaker’s opening, like in the exercise, directly affects the total area available for sound to radiate, playing a critical role in determining the overall intensity of the sound produced.
Electrical to Sound Power Conversion
Electrical-to-sound power conversion is the process by which electrical energy is transformed into sound energy by devices like loudspeakers. This process is not entirely efficient, meaning not all electrical power input is converted into sound power. Some of the energy is lost as heat or through mechanical inefficiencies.

To determine how efficiently a loudspeaker converts electrical power into sound power, you can calculate its efficiency in percentage terms:
  • Efficiency = (Sound Power / Electrical Power) × 100%
In the provided exercise, we calculated that the loudspeaker has about 1.985% efficiency. This means only a small fraction of the power supplied is actually converted into sound. This efficiency level, which seems low, is typical for loudspeakers because converting all electrical energy to sound without any losses is nearly impossible.

Understanding the efficiency of this conversion process is significant, especially in designing better audio equipment and systems, ensuring that energy use is optimized and devices function effectively within their constraints.

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Most popular questions from this chapter

A listener doubles his distance from a source that emits sound uniformly in all directions. There are no reflections. By how many decibels does the sound intensity level change?

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

To navigate, a porpoise emits a sound wave that has a wavelength of \(1.5 \mathrm{cm} .\) The speed at which the wave travels in seawater is \(1522 \mathrm{m} / \mathrm{s} .\) Find the period of the wave.

(a) A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed \(v\) of the wave on the rope in terms of the distance \(y\) above the bottom end of the rope and the magnitude \(g\) of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{m}\) and \(2.0 \mathrm{m}\) above the bottom end of the rope.

A monatomic ideal gas \((\gamma=1.67)\) is contained within a box whose volume is \(2.5 \mathrm{m}^{3} .\) The pressure of the gas is \(3.5 \times 10^{s} \mathrm{Pa}\). The total mass of the gas is \(2.3 \mathrm{kg}\). Find the speed of sound in the gas.

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