/*! 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 27 A certain sound source is increa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A certain sound source is increased in sound level by \(30.0 \mathrm{~dB}\). By what multiple is (a) its intensity increased and (b) its pressure amplitude increased?

Short Answer

Expert verified
Intensity increases by 1000 times; pressure amplitude increases by about 31.62 times.

Step by step solution

01

Understanding Decibels and Intensity

Decibels (dB) measure the sound intensity level on a logarithmic scale. The formula for sound intensity level is given by: \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \), where \( L \) is the level in decibels, \( I \) is the intensity, and \( I_0 \) is the reference intensity. Here, the increase is 30 dB.
02

Relating Intensity Change to Decibels

To find the change in intensity, we use the formula from the previous step: \( 30 = 10 \log_{10}\left(\frac{I'}{I}\right) \), where \( I' \) is the new intensity. Solving for \( \frac{I'}{I} \), we have \( 3 = \log_{10}\left(\frac{I'}{I}\right) \). By exponentiating both sides, \( \frac{I'}{I} = 10^3 = 1000 \). Thus, intensity increases by a factor of 1000.
03

Understanding Pressure Amplitude and Intensity

Intensity is proportional to the square of the pressure amplitude \( \Delta P \), which gives us: \( I \propto (\Delta P)^2 \). This means \( \frac{I'}{I} = \left(\frac{\Delta P'}{\Delta P}\right)^2 \).
04

Calculating the Change in Pressure Amplitude

Since \( \frac{I'}{I} = 1000 \), we have \( 1000 = \left(\frac{\Delta P'}{\Delta P}\right)^2 \). Taking the square root of both sides gives \( \frac{\Delta P'}{\Delta P} = \sqrt{1000} \approx 31.62 \). Thus, the pressure amplitude increases by a factor of approximately 31.62.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Decibels
Decibels (dB) are the unit used to measure sound intensity level. They are particularly useful because they use a logarithmic scale to represent the intensity. This helps manage very large variations in intensity levels that the human ear can hear. As a result, a small change in decibels can represent a large change in intensity.

The formula to calculate the sound intensity level in decibels is \[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]where:
  • \( L \) is the sound level in decibels
  • \( I \) is the sound intensity
  • \( I_0 \) is the reference intensity (typically \( 10^{-12} \text{W/m}^2 \) for air)
The logarithmic nature of decibels means an increase of 10 dB actually represents a tenfold increase in intensity. This is why changes in dB levels can turn into significant intensity changes.
Intensity Increase
Sound intensity refers to the amount of sound energy passing through a unit area. When you increase sound level by a given amount in decibels, this corresponds to an exponential increase in sound intensity due to the logarithmic base of dB calculations.

For example, an increase of 30 dB corresponds to the process explained in the step-by-step solution:
  • Using the formula \( 30 = 10 \log_{10}\left(\frac{I'}{I}\right) \) implies a tenfold increase in sound intensity for every 10 dB.
  • Solving this equation shows that the new intensity \( I' \) is 1000 times the original intensity \( I \). Thus, the intensity is increased by a factor of 1000.
Essentially, the larger the decibel increase, the more dramatic the increase in the actual energy or intensity of the sound.
Pressure Amplitude
Pressure amplitude in the context of sound refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In simple terms, it describes how "loud" the sound wave is perceived.

The relationship between intensity and pressure amplitude is expressed as:
  • Sound intensity \( I \) is proportional to the square of the pressure amplitude \( (\Delta P)^2 \).
  • This means if the intensity increases, the pressure amplitude must also increase, but not linearly.
From the initial problem, when intensity increases by a factor of 1000, the formula used is:\[ \frac{I'}{I} = \left(\frac{\Delta P'}{\Delta P}\right)^2 \]Solving this equation confirms that the pressure amplitude is increased by approximately 31.62 times. This is because the change in pressure amplitude follows a square root function due to its proportional relationship with intensity.
Logarithmic Scale
The logarithmic scale is the mathematical foundation used in measuring sound levels with decibels. It allows us to represent very large ranges of values in a more manageable and compressed form. Rather than a linear scale, which can be unwieldy when describing phenomena like sound, a logarithmic scale interprets exponential relationships more efficiently.

In the context of sound intensity:
  • Increases in sound intensity levels correspond to increases in decibels, which are calculated using the base-10 logarithm.
  • Each 10 dB increase represents a tenfold increase in intensity, reflecting the logarithmic nature.
This scale is not just applied in sound, but in many fields where quantities can span several orders of magnitude, e.g., pH levels in chemistry or Richter scale for earthquakes. By using a logarithmic scale, we gain a clearer understanding of how changes in numeric values relate to perceived changes in sound or physical effects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose a spherical loudspeaker emits sound isotropically at \(10 \mathrm{~W}\) into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance \(d=3.0 \mathrm{~m}\) from the center of the source? (b) What is the ratio of the wave amplitude at \(d=4.0 \mathrm{~m}\) to that at \(d=3.0 \mathrm{~m}\) ?

Two identical piano wires have a fundamental frequency of \(600 \mathrm{~Hz}\) when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of \(6.0\) beats/s when both wires oscillate simultaneously?

A point source emits \(30.0\) W of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 200 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement, for a different sinusoidal sound wave, air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?

Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength. Rather, they change the value of \(d V / d p-\) that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If \(f_{s} / f_{i}=0.333\), what is the ratio \((d V / d p)_{s} /(d V / d p)_{i} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.