/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 You are trying to overhear a jui... [FREE SOLUTION] | 91Ó°ÊÓ

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You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?

Short Answer

Expert verified
Move to 0.15 meters away for the sound level to be 60 dB.

Step by step solution

01

Convert Decibels to Intensity Ratio

The sound level in decibels (dB) is a logarithmic measure of the ratio of the intensity of a sound wave to a reference intensity. We can use the formula \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \), where \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity (\( 10^{-12} \text{ W/m}^2 \)). We first calculate the intensity ratio for both 20.0 dB and 60.0 dB.
02

Calculate Intensities at Both Distances

For 20 dB: \( 20 = 10 \cdot \log_{10} \left( \frac{I_1}{I_0} \right) \). Solving gives \( \frac{I_1}{I_0} = 10^2 = 100 \); hence, \( I_1 = 100I_0 \).For 60 dB: \( 60 = 10 \cdot \log_{10} \left( \frac{I_2}{I_0} \right) \). Solving gives \( \frac{I_2}{I_0} = 10^6 = 1,000,000 \); hence, \( I_2 = 1,000,000I_0 \).
03

Use Inverse Square Law

Sound intensity is inversely proportional to the square of the distance from the source: \( I \propto \frac{1}{r^2} \). Given \( I_1 = 100I_0 \) at 15.0 m, and we want \( I_2 = 1,000,000I_0 \), we use the relationship \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \).
04

Solve for the New Distance

Plug the values into the inverse square relationship: \( \frac{100I_0}{1,000,000I_0} = \left( \frac{r_2}{15} \right)^2 \). This simplifies to \( \frac{1}{10,000} = \left( \frac{r_2}{15} \right)^2 \). Solving for \( r_2 \), we get \( r_2 = 15 \times \frac{1}{100} = 0.15 \text{ meters} \).

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Key Concepts

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

Decibel Scale
Sound levels are measured using the decibel (dB) scale, which is a logarithmic way to describe a ratio. The decibel scale helps us understand how loud a sound is compared to a reference sound. The reference intensity, usually denoted as \( I_0 \), is set at the threshold of hearing, which is \( 10^{-12} \text{ W/m}^2 \). This threshold represents the quietest sound that a healthy human ear can detect.
To calculate the intensity level of a sound in decibels, we use the formula:
  • \( \beta = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \)
Here, \( \beta \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity. This formula shows that every increase of 10 dB represents a tenfold increase in intensity. For example, a sound that is 60 dB is not just three times louder than a 20 dB sound, but rather 10,000 times more intense. This logarithmic nature is why smaller dB changes represent larger changes in intensity, which is crucial in understanding sound dynamics.
Inverse Square Law
The inverse square law is a fundamental principle that describes how sound intensity decreases with distance. According to this law, the intensity of sound is inversely proportional to the square of the distance from the sound source. Mathematically, we express this relationship as:
  • \( I \propto \frac{1}{r^2} \)
Here, \( I \) denotes the intensity of the sound, while \( r \) represents the distance from the sound source. This means that when the distance is doubled, the sound intensity becomes one-fourth as intense. In the context of our exercise, moving from 15 meters to a closer distance significantly increases the sound intensity.To solve such problems, we typically set up a ratio using the intensities at two different distances. If \( I_1 \) is the intensity at the first distance \( r_1 \), and \( I_2 \) is the intensity at the desired distance \( r_2 \), then:
  • \( \frac{I_1}{I_2} = \left( \frac{r_2}{r_1} \right)^2 \)
Solving this equation will help you find how close you need to move to hear a sound more clearly.
Sound Intensity
Sound intensity is a measure of the power carried by sound waves per unit area, usually expressed in watts per square meter (\( \text{W/m}^2 \)). It indicates how much sound energy passes through a particular area and is a key factor in determining how loud a sound is perceived by a listener.
In our example, we are given two sound levels: 20 dB and 60 dB. To find the sound intensity at these levels, we first use the decibel formula:
  • At 20 dB: \( I_1 = 100 I_0 \)
  • At 60 dB: \( I_2 = 1,000,000 I_0 \)
These calculations show the vast difference in sound intensity between the two levels. Understanding the sound intensity allows us to apply the inverse square law effectively. By knowing the sound intensity at a starting distance, you can find out how close you need to be to a sound source to experience a desired intensity. This essential concept highlights the relationship between intensity, distance, and perceived loudness, which all stem from the principles of physics governing sound.

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Most popular questions from this chapter

A 2.00-MHz sound wave travels through a pregnant woman's abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 72 beats per second are detected. The speed of sound in body tissue is 1500 m/s. Calculate the speed of the fetal heart wall at the instant this measurement is made.

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of 3 Hz, which increases to 4 Hz when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3-Hz beats? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 cm. Find the beat frequency that they produce when playing together in their fundamentals.

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750 Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?

In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce standing waves. This is of concern because the large pressure amplitude in an antinode can damage tissues. For a frequency of 1.0 MHz, what is the distance between antinodes in tissue? (a) 0.38 mm; (b) 0.75 mm; (c) 1.5 mm; (d) 3.0 mm.

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