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Question: A sound-insulating door reduces the sound level by\({\bf{30}}\,{\bf{dB}}\). What fraction of the sound intensity passes through this door?

Short Answer

Expert verified

The \(\frac{1}{{1000}}\) sound intensity passes through the door.

Step by step solution

01

Concept

The equation of sound intensity is,

\(\beta = 10\log \frac{I}{{{I_0}}}\)

The intensity of the emitted sound is\(I\)and the intensity of the threshold audible sound is\({I_0} = {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

02

Given Data

The sound decreases by \(\beta = - 30\;{\rm{dB}}\).

03

Calculation

The intensity decreases as it passes through the door. So you can write,

\(\begin{array}{c}\beta = 10\log \frac{I}{{{I_0}}} = - 30\\\log \frac{I}{{{I_0}}} = - 3\\\frac{I}{{{I_0}}} = {10^{ - 3}}\\\frac{I}{{{I_0}}} = \frac{1}{{1000}}\end{array}\)

Hence, the\(\frac{1}{{1000}}\) sound intensity passes through the door.

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Most popular questions from this chapter

Question: The 鈥渁lpenhorn鈥 (Fig. 12鈥42) was once used to send signals from one Alpine village to another. Since lower frequency sounds are less susceptible to intensity loss, long horns were used to create deep sounds. When played as a musical instrument, the alpenhorn must be blown in such a way that only one of the overtones is resonating. The most popular alpenhorn is about \({\bf{3}}{\bf{.4}}\,{\bf{m}}\) long, and it is called the \({{\bf{F}}^{\bf{\# }}}\) horn. What is the fundamental frequency of this horn, and which overtone is close to\({{\bf{F}}^{\bf{\# }}}\)? (See Table12鈥3.) Model as a tube open at both ends.

A uniform narrow tube \({\bf{1}}{\bf{.70}}\;{\bf{m}}\) long is open at both ends. It resonates at two successive harmonics of frequencies \({\bf{275}}\;{\bf{Hz}}\) and \({\bf{330}}\;{\bf{Hz}}\) . What is (a) the fundamental frequency, and (b) the speed of sound in the gas in the tube?

Question: (I) A piano tuner hears one beat every 2.0 s when trying to adjust two strings, one of which is sounding 350 Hz. How far off in frequency is the other string?

A guitar string vibrates at a frequency of 330 Hz with a wavelength 1.40 m. The frequency and wavelength of this sound in air (20掳C) as it reaches our ears is

(a) same frequency, same wavelength.

(b) higher frequency, same wavelength.

(c) lower frequency, same wavelength.

(d) same frequency, longer wavelength.

(e) same frequency, shorter wavelength.

Why are the frets on a guitar (Fig. 12鈥30) spaced closer together as you move up the fingerboard toward the bridge?

FIGURE 12鈥30 Question 9.

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