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Assume that a loudspeaker broadcasts sound equally in all directions and produces sound with a level of 103 \(\mathrm{dB}\) at a distance of 1.60 \(\mathrm{m}\) from its center. (a) Find its sound power output. (b) If the salesperson claims to be giving you 150 \(\mathrm{W}\) per channel, he is referring to the electrical power input to the speaker. Find the efficiency of the speaker-that is, the fraction of input power that is converted into useful output power.

Short Answer

Expert verified
The sound power output is calculated using intensity and area, and the efficiency is determined by comparing the sound power output with the electrical power input.

Step by step solution

01

Calculate the intensity at 1.60 m

Use the formula for sound intensity level in decibels (dB): \( I = (10^{-12} \text{ W/m}^2) \times 10^{(L/10)} \), where \( I \) is the sound intensity and \( L \) is the sound level in dB. Plug in \( L = 103 \) dB to find \( I \).
02

Calculate the power output of the loudspeaker

The sound power output can be found by using the formula \( P = I \times A \), where \( P \) is the power, \( I \) is the sound intensity, and \( A \) is the surface area of the sphere with radius equal to the given distance (1.60 m). Use \( A = 4\pi r^2 \) to calculate the area of the sphere.
03

Calculate the efficiency of the speaker

Efficiency (\( \eta \)) can be calculated using the formula \( \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\% \). Use the power output from Step 2 and the given input power (150 W) to find the efficiency.

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Key Concepts

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

Decibels
The decibel (dB) is a unit of measurement that reflects the relative loudness of sounds. It is a logarithmic unit, which means it helps us compare sounds that have a large range of intensities in a more manageable way. The formula to determine the sound intensity level in decibels is given by:
\( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \),
where \( L \) is the sound level in decibels, \( I \) is the sound intensity in watts per square meter (W/m\textsuperscript{2}), and \( I_0 = 10^{-12} \text{W/m}^2 \) is the reference sound intensity, considered the threshold of human hearing.

In practical terms, the sound intensity level allows us to understand how loud a sound is. If a loudspeaker is measured at 103 dB at a certain distance, this reflects its loudness relative to the reference intensity. For every 10 dB increase in sound level, the perceived loudness of the sound roughly doubles. This is crucial for quantifying the output of devices like loudspeakers and for ensuring that sound levels are within safe listening ranges.
Sound Power Output
Sound power output is a measure of the acoustic energy emitted by a source per unit time, measured in watts (W). It represents the total energy that a sound source, such as a loudspeaker, can distribute in all directions. The sound intensity (\( I \) - measured in W/m\textsuperscript{2}) at a specific distance from the source is used to calculate this power output. The formula to find the power output of a sound source is:
\( P = I \times A \),
where \( P \) is the sound power, \( A \) is the area through which the sound is spreading, and \( I \) is the sound intensity at that area. For a source radiating sound equally in all directions (isotropically), \( A \) is the surface area of a sphere with radius equal to the distance from the source (4\pi r^2).

To determine the loudspeaker's sound power output, you would calculate the area over which the sound is distributed at the given distance and multiply it by the sound intensity at that distance. For our exercise, the radius would be 1.60 m, so you would use it to find the spherical area and thus calculate the sound power output. Understanding sound power output is valuable for designing audio equipment and for controlling noise pollution in various environments.
Loudspeaker Efficiency
Loudspeaker efficiency is the ratio of the acoustic power output to the electrical power input, often expressed as a percentage. It gives us an indication of how effectively a loudspeaker converts the electrical energy it receives into sound. The formula used to calculate the efficiency of a loudspeaker is:
\( \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\% \),
where \( \eta \) is the efficiency, \( P_{\text{output}} \) is the acoustic power output, and \( P_{\text{input}} \) is the electrical power provided to the loudspeaker.

In practical applications, loudspeakers are not 100% efficient; a significant portion of the input power is typically converted into heat rather than sound. For example, if we use the output power calculated previously and the claimed 150 W per channel of input power, we can determine the efficiency of the loudspeaker. This efficiency is critical to understand when designing sound systems, as higher efficiency can reduce power consumption and the need for larger amplifiers while still achieving the desired loudness.

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