/*! 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 83 A microphone receiving a pure so... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 83

A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally \(2.00 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2},\) but is turned up until the amplitude increases by \(30.0 \%,\) what is the new intensity?

Short Answer

Expert verified
The new intensity of the sound after increasing the amplitude by 30% is \(3.38 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\).

Step by step solution

01

Finding the new amplitude

The given sound intensity is \(2.00 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\). The amplitude is increased by 30%. So let the original amplitude be A, then the new amplitude (A') can be calculated as follows: A' = A + (A x 0.30) Since we're only interested in the intensity after the amplitude increase, we only need the proportionality between the original amplitude and the new amplitude squares. Ratio = \(\frac{A'^{2}}{A^{2}}\)
02

Calculate the ratio of amplitude squares

Now, substitute the new amplitude in terms of the original amplitude (A) into the ratio formula: Ratio = \(\frac{(A + 0.30A)^{2}}{A^{2}} = \frac{(1.30A)^{2}}{A^{2}}\) Ratio = \((1.30)^{2} = 1.69\)
03

Calculate the new intensity

Now that we have the proportionality ratio between the original amplitude and the new amplitude squares, we can calculate the new intensity (I'): I' = Ratio 脳 original intensity I' = 1.69 脳 \(2.00 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\) I' = \(3.38 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\) The new intensity of the sound after the amplitude is increased by 30% is \(3.38 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\).

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Key Concepts

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

Sound Amplitude
Sound amplitude is a measure of how intense a sound wave is. It indicates the maximum displacement of particles in the medium through which the sound wave travels. For sound waves, higher amplitude means a louder sound. Sound amplitude is directly related to the energy and intensity of the wave; therefore, increasing the amplitude of a wave increases its intensity. In practical terms, sound amplitude can be visualized as the "height" of the wave on an oscilloscope screen. A higher wave corresponds to a greater amplitude or louder sound. In the context of the problem, an increase of 30% in amplitude signifies that the sound intensity will change as well.
Oscilloscope
An oscilloscope is an electronic instrument used to visualize waveforms, making it a valuable tool for observing sound waves. It displays the voltage versus time graph of a wave, which allows us to see key characteristics like amplitude and frequency. When a microphone converts sound waves into electrical signals, an oscilloscope can then show these converted signals as a waveform on its screen.
  • Amplitude appears as the peak height of the wave.
  • The time between wave peaks helps us determine frequency.
  • Changes in amplitude affect the overall shape and intensity of the wave seen on the screen.
By using an oscilloscope, we can observe that amplifying a sound wave causes the peaks on the screen to increase in height, corresponding to the increased amplitude and intensity of the sound.
Wave Intensity Calculation
Wave intensity in physics refers to the power transmitted per unit area in a direction perpendicular to that area. For sound waves, intensity is dependent on both the amplitude and frequency of the waves. Generally, the intensity of a sound wave is calculated using the proportionality of amplitude. In such calculations, it is essential to understand the relationship: intensity is proportional to the square of the amplitude.

In the given exercise, after increasing the amplitude by 30%, the new intensity can be derived using the formula: \[I' = \left(\frac{A'}{A}\right)^2 \times I\].
Here, \( A' = 1.30 \times A \), which leads to: \(I' = (1.30)^2 \times 2.00 \times 10^{-5} \mathrm{W}/\mathrm{m}^2 \). This gives the new intensity \(I'\) as \(3.38 \times 10^{-5} \mathrm{W}/\mathrm{m}^2\).

Remember, in sound wave calculations, a small change in amplitude can cause a significant change in intensity due to the squared relationship.

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Most popular questions from this chapter

A string of length \(5 \mathrm{m}\) and a mass of \(90 \mathrm{g}\) is held under a tension of 100 N. A wave travels down the string that \(\quad\) is \(\quad\) modeled \(\quad\) as \(y(x, t)=0.01 \mathrm{m} \sin \left(0.40 \mathrm{m}^{-1} x-1170.12 \mathrm{s}^{-1}\right) .\) What is the power over one wavelength?

A Girl Scout is taking a 10.00-km hike to earn a merit badge. While on the hike, she sees a cliff some distance away. She wishes to estimate the time required to walk to the cliff. She knows that the speed of sound is approximately 343 meters per second. She yells and finds that the echo returns after approximately 2.00 seconds. If she can hike \(1.00 \mathrm{km}\) in 10 minutes, how long would it take her to reach the cliff?

A transverse wave on a string is modeled with the wave function \(y(x, t)=0.10 \mathrm{m} \sin \left(0.15 \mathrm{m}^{-1} x+1.50 \mathrm{s}^{-1} t+0.20\right)\) Find the wave velocity. (b) Find the position in the \(y\) -direction, the velocity perpendicular to the motion of the wave, and the acceleration perpendicular to the motion of the wave, of a small segment of the string centered at \(x=0.40 \mathrm{m}\) at time \(t=5.00 \mathrm{s}\).

Consider two wave functions \(y(x, t)=0.30 \mathrm{cm} \sin \left(3 \mathrm{m}^{-1} x-4 \mathrm{s}^{-1} t\right)\) \(y(x, t)=0.30 \mathrm{cm} \sin \left(3 \mathrm{m}^{-1} x+4 \mathrm{s}^{-1} t\right) .\) Write a wave function for the resulting standing wave.

A device called an insolation meter is used to measure the intensity of sunlight. It has an area of \(100 \mathrm{cm}^{2}\) and registers \(6.50 \mathrm{W}\). What is the intensity in \(\mathrm{W} / \mathrm{m}^{2}\) ?
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