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Chapter 17: Problem 76

Find the ratios (greater to smaller) of the (a) intensities, (b) pressure amplitudes, and (c) particle displacement amplitudes for two sounds whose sound levels differ by \(37 \mathrm{~dB}\).

Short Answer

Expert verified
Intensity ratio: 5012; Pressure and displacement amplitude ratios: 70.8.

Step by step solution

01

Introduction to Sound Levels

Sound levels in decibels (dB) measure the intensity of sounds on a logarithmic scale. If two sounds differ by a certain number of decibels, this corresponds to a ratio of their intensities.
02

Intensity Ratio Formula

The formula to find the ratio of intensities given a decibel difference is:\[ I_1 / I_2 = 10^{( ext{dB difference}/10)} \] We apply this formula for a 37 dB difference.
03

Calculate Intensity Ratio

Substituting the given difference into the formula:\[ I_1 / I_2 = 10^{(37/10)} = 10^{3.7} \] Calculating this value gives approximately 5012.
04

Relation between Pressure Amplitude and Intensity

The intensity of a sound is proportional to the square of the pressure amplitude. Therefore, the pressure amplitude ratio is the square root of the intensity ratio. Thus,\[ P_1 / P_2 = \sqrt{I_1 / I_2} \]
05

Calculate Pressure Amplitude Ratio

Substituting the intensity ratio of 5012 into the formula gives:\[ P_1 / P_2 = \sqrt{5012} \] This value is approximately 70.8.
06

Relation between Particle Displacement Amplitude and Pressure Amplitude

The pressure amplitude is proportional to the displacement amplitude; hence they have the same ratio. Therefore, the amplitude ratio is the same as the pressure amplitude ratio.
07

Calculate Displacement Amplitude Ratio

The particle displacement amplitude ratio is the same as the pressure amplitude ratio, which is approximately 70.8.

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Key Concepts

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

Decibel Scale
Sound is an integral part of our world, and measuring it accurately is important for various applications. One of the key concepts in sound measurement is the decibel scale. The decibel (dB) scale is a logarithmic way of representing sound intensity levels. This means rather than counting up in regular units, each step on the dB scale represents a tenfold increase or decrease in intensity.

Because the human ear can detect a phenomenal range of sound intensities鈥攆rom the faint rustle of leaves to the roar of a jet engine鈥攁 linear scale would be impractical. Instead, the logarithmic nature of the dB scale allows us to condense this wide range into manageable numbers. For instance, if two sounds differ by 37 dB, they differ significantly in terms of energy. On a linear scale, a 37 dB difference translates to an intensity ratio of:
  • \[I_1 / I_2 = 10^{(37/10)} = 10^{3.7} \]
This calculation results in a difference of about 5012 times, indicating the more intense sound has much more energy.
Pressure Amplitude
Pressure amplitude is essentially the variation in air pressure caused by sound waves. It is closely linked to how humans perceive the loudness of sound. When we speak of sound intensity in relation to pressure amplitude, it's important to remember that sound intensity is proportional to the square of the pressure amplitude.

This relationship means that if two sounds differ in sound intensity, their pressure amplitude ratio can be calculated by taking the square root of the intensity ratio.
  • \[P_1 / P_2 = \sqrt{I_1 / I_2} = \sqrt{5012}\approx 70.8 \]
So, the pressure amplitude of one sound is approximately 70.8 times larger than the other when there鈥檚 a 37 dB sound level difference.
Particle Displacement
In sound waves, particle displacement refers to how far individual particles within a medium move from their rest position as a sound wave passes. This is different from pressure amplitude, which deals with variations in pressure.

Despite this difference, particle displacement shares a direct proportionality with pressure amplitude for a given sound. Therefore, when calculating their ratios, you use the same value obtained from the pressure amplitude ratio calculation. This means that for two sounds whose sound levels differ by 37 dB, the particle displacement amplitude ratio is also approximately 70.8.
  • Practical importance: Understanding particle displacement is crucial for sound engineering and noise control.
  • Different mediums: The displacement can vary significantly based on the medium (air, water, etc.) the sound travels through.

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Most popular questions from this chapter

A detector initially moves at constant velocity directly toward a stationary sound source and then (after passing it) directly from it. The emitted frequency is \(f\). During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\). If the frequencies are related by \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=0.500,\) what is the ratio \(v_{D} / v\) of the speed of the detector to the speed of sound?

You can estimate your distance from a lightning stroke by counting the seconds between the flash you see and the thunder you later hear. By what integer should you divide the number of seconds to get the distance in kilometers?

When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation," produces a sound pulse \(-\) the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside. If the pulse has a sound level of \(62 \mathrm{~dB}\) at your ear, estimate the rate at which energy is produced by the cavitation.

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) over takes and passes a cyclist pedaling a bike at \(2.44 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz}\). How fast is the ambulance moving?

Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous otoacoustic emission is rarely aware of the sound, except perhaps in a noisefree environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of \(1665 \mathrm{~Hz}\) and a pressure amplitude of \(1.13 \times 10^{-3} \mathrm{~Pa}\). What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?
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