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(a) If two sounds differ by \(5.00 \mathrm{~dB}\), find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Short Answer

Expert verified
The ratio of the intensity of the louder sound to that of the softer is approximately \(10^{0.5}\). The sound intensity levels differ by \(20dB\). The sound intensity level increases approximately by \(10 \log(2)dB\) when the intensity doubles.

Step by step solution

01

Calculating the ratio of the intensities of two sounds

We have \(L_1 - L_2 = 5dB\), implying \(10 \log(I_1/I_0) - 10 \log(I_2/I_0) = 5\). This can be rewritten as \(10 \log(I_1/I_2) = 5\), leading to \(\log(I_1/I_2) = 0.5\). Taking the exponential of both sides to remove the logarithm, we have \(I_1 / I_2 = 10^{0.5}\).
02

Finding the difference in their sound intensity levels

We have \(I_1 = 100I_2\), implying \(10 \log(I_1/I_0) = 10 \log(100 I_2/I_0)\). This can be written as \(10 \log(I_1/I_2) = 10 \log(100)\), simplifying to \(\log(I_1/I_2) = 2\). Taking the logarithm base 10 and multiplying by 10 gives \(L_1 - L_2 = 20dB\).
03

Calculating the increase in sound intensity level

We have \(I_1 = 2I_2\), leading to \(10 \log(I_1/I_0) = 10 \log(2I_2/I_0)\). This can be simplified as \(10 \log(I_1/I_2) = 10 \log(2)\), leading to \(L_1 - L_2 = 10 \log(2)dB\).

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Key Concepts

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

Decibels
Decibels (dB) are units used to measure sound intensity. They express sound levels in a logarithmic scale rather than a linear scale. By using a logarithmic scale, the wide range of sound intensities present in our environment can be handled in a more manageable way.
Sound intensity measured in decibels represents the power per unit area. If you're familiar with the concept of noise levels, the decibel scale makes it easier to quantify how intense a sound actually is. A slight increase in decibels signifies a large increase in intensity; for example, a sound that is 10 decibels higher is actually 10 times more intense.
  • 0 dB is the threshold of hearing.
  • 30 dB is a whisper, 60-70 dB is typical conversation.
  • 120 dB is the threshold of pain.
In essence, the decibel scale is useful because it allows us to express very large or small numbers in a compact form, which is easier to communicate and work with especially in acoustics.
Sound Intensity Level
Sound intensity level is the measure of sound power that is transferred through a certain area in space. Unlike sound intensity, which measures the force per unit area, intensity level specifically refers to how loud a sound is perceived in decibels.
The sound intensity level is given by the formula:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where \(L\) is the sound level in decibels, \(I\) is the sound intensity, and \(I_0\) is the reference intensity (usually set at the threshold of hearing, \(10^{-12} \text{W/m}^2\)).
To work out the difference in sound intensity levels when the intensity changes, you subtract the two sound levels. This often involves using the properties of logarithms, since these calculations are done on a logarithmic scale.
  • A 10 dB increase corresponds to a tenfold increase in intensity.
  • A 20 dB increase implies a hundredfold increase in intensity.
Thus, when sound levels change, it's important to remember that a small change in decibels implies a large change in actual intensity.
Logarithms
Logarithms are mathematical operations that are used to scale numbers, making them easier to understand and work with. They are the inverse operation of exponentiation, meaning \( \log_b(x) = y \) if and only if \(b^y = x\).
In sound intensity calculations, base 10 logarithms are particularly useful. The decibel scale operates on these base 10 logarithms. For example, when comparing two sound intensities \(I_1\) and \(I_2\), we use:
\[ \log \left( \frac{I_1}{I_2} \right) \]
The benefit of logarithmic operations in acoustics is that they transform multiplicative relationships into additive ones. This is crucial when working with sound, as it allows us to use manageable numbers to express vast differences in intensities.
  • Logarithms can simplify complex multiplicative problems.
  • They reduce large ranges of data into more consumable figures.
Overall, understanding logarithms helps us in comprehending the decibel scale and in performing sound intensity calculations easily.

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Most popular questions from this chapter

You blow across the open mouth of an empty test tube and produce the fundamental standing wave in the \(14.0-\mathrm{cm}\) -long air column in the test tube, which acts as a stopped pipe. (a) What is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

A police siren of frequency \(f_{\text {siren }}\) is attached to a vibrating platform. The platform and siren oscillate up and down in simple harmonic motion with amplitude \(A_{\mathrm{p}}\) and frequency \(f_{\mathrm{p}}\). (a) Find the maximum and minimum sound frequencies that you would hear at a position directly above the siren. (b) At what point in the motion of the platform is the maximum frequency heard? The minimum frequency? Explain.

(a) A sound source producing \(1.00 \mathrm{kHz}\) waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

A long spring such as a Slinkyrs is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke's law has mass \(m,\) length \(L\), and force constant \(k^{\prime}\), the speed of longitudinal waves on the spring is \(v=L \sqrt{k^{\prime} / m}\) (see Section 16.2 ). (b) Evaluate \(v\) for a spring with \(m=0.250 \mathrm{~kg}, L=2.00 \mathrm{~m},\) and \(k^{\prime}=1.50 \mathrm{~N} / \mathrm{m}\)

(a) In a liquid with density \(1300 \mathrm{~kg} / \mathrm{m}^{3},\) longitudinal waves with frequency \(400 \mathrm{~Hz}\) are found to have wavelength \(8.00 \mathrm{~m}\). Calculate the bulk modulus of the liquid. (b) A metal bar with a length of \(1.50 \mathrm{~m}\) has density \(6400 \mathrm{~kg} / \mathrm{m}^{3}\). Longitudinal sound waves take \(3.90 \times 10^{-4} \mathrm{~s}\) to travel from one end of the bar to the other. What is Young's modulus for this metal?

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