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Two loudspeakers, \(A\) and \(B,\) radiate sound uniformly in all directions in air at \(20^{\circ} \mathrm{C}\) . The acoustic power output from \(A\) is \(8.00 \times 10^{-4} \mathrm{W}\) , and from \(B\) it is \(6.00 \times 10^{-5} \mathrm{W}\) . Both loudspeakcrs are vibrating in phase at a frequency of 172 \(\mathrm{Hz}\) (a) Determine the difference in phase of the two signals at a point \(C\) along the line joining \(A\) and \(B, 3.00 \mathrm{m}\) from \(B\) and 4.00 \(\mathrm{m}\) from \(A(\mathrm{Fig} .16 .46)\) . (b) Determine the intensity and sound intensity level at \(C\) from speaker \(A\) if speaker \(B\) is turned off and the intensity and sound intensity level at point \(C\) from speaker \(B\) if speaker \(A\) is turned off. (c) With both speakers on, what are the intensity and sound intensity level at \(C ?\)

Short Answer

Expert verified
a) Phase difference is approximately 3.144 radians. b) Intensities: 3.98脳10鈦烩伓 W/m虏 (95.99 dB), 5.31脳10鈦烩伓 W/m虏 (97.25 dB). c) Total intensity: 2.76脳10鈦烩伓 W/m虏 (94.41 dB).

Step by step solution

01

Determine the Speed of Sound

At an air temperature of \(20^{\circ} \mathrm{C}\), the speed of sound \(v\) is approximately \(343 \, \mathrm{m/s}\). This is a commonly-used speed for calculations involving sound in air at room temperature.
02

Calculate Wavelength

The wavelength \(\lambda\) of the sound waves can be calculated using the formula \(\lambda = \frac{v}{f}\), where \(v\) is the speed of sound and \(f\) is the frequency. Here, \(\lambda = \frac{343 \, \mathrm{m/s}}{172 \, \mathrm{Hz}} = 1.9953 \, \mathrm{m}\).
03

Determine Phase Difference

The phase difference \(\Delta \phi\) between two points can be found using \(\Delta \phi = \frac{2\pi (d_B - d_A)}{\lambda}\), where \(d_A\) and \(d_B\) are the distances from speakers \(A\) and \(B\) to point \(C\), respectively. Substituting the given values, \(\Delta \phi = \frac{2\pi (4.00 \, \mathrm{m} - 3.00 \, \mathrm{m})}{1.9953 \, \mathrm{m}} \approx 3.144\) radians.
04

Calculate Intensity at C from Speaker A

Intensity \(I\) is given by \(I = \frac{P}{4\pi r^2}\), where \(P\) is the power output and \(r\) is the distance from the source. For Speaker \(A\), \(I_A = \frac{8.00 \times 10^{-4} \, \mathrm{W}}{4\pi (4.00 \, \mathrm{m})^2} \approx 3.98 \times 10^{-6} \, \mathrm{W/m}^2\).
05

Calculate Sound Intensity Level from Speaker A

Sound intensity level \(L\) in decibels \(\mathrm{dB}\) is calculated as \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\), where \(I_0 = 10^{-12} \, \mathrm{W/m}^2\). So, \(L_A = 10 \log_{10} \left( \frac{3.98 \times 10^{-6}}{10^{-12}} \right) \approx 95.99 \, \mathrm{dB}\).
06

Calculate Intensity at C from Speaker B

Similarly, for Speaker \(B\), \(I_B = \frac{6.00 \times 10^{-5} \, \mathrm{W}}{4\pi (3.00 \, \mathrm{m})^2} \approx 5.31 \times 10^{-6} \, \mathrm{W/m}^2\).
07

Calculate Sound Intensity Level from Speaker B

For Speaker \(B\), the sound intensity level \(L_B = 10 \log_{10} \left( \frac{5.31 \times 10^{-6}}{10^{-12}} \right) \approx 97.25 \, \mathrm{dB}\).
08

Calculate Total Intensity with Both Speakers On

When both speakers are on, the total intensity is the sum of the individual intensities, modified by the interference term, given as \(I = I_A + I_B + 2\sqrt{I_A I_B}\cos(\Delta \phi)\). Using \(\Delta \phi = 3.144\) radians, which is approximately \(\pi\), \(\cos(\Delta \phi) = -1\), thus \(I \approx I_A + I_B - 2\sqrt{I_A I_B} \approx 2.76 \times 10^{-6} \, \mathrm{W/m}^2\).
09

Calculate Total Sound Intensity Level at C

The total sound intensity level is \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\), so \(L \approx 10 \log_{10} \left( \frac{2.76 \times 10^{-6}}{10^{-12}} \right) \approx 94.41 \, \mathrm{dB}\).

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through a medium like air, water, or solids. They're created when an object vibrates, sending waves through the medium. In the case of loudspeakers, they convert electrical signals into sound waves that we can hear.

Sound waves are longitudinal waves, meaning they move in the same direction as the vibration of the particles in the medium. Imagine a slinky being pushed and pulled; that's how particles in the medium move back and forth to create sound waves. The speed of sound in air, affected by temperature and pressure, is approximately 343 m/s at 20掳C. This speed is crucial for determining other properties like wavelength and frequency.

Sound waves have different properties:
  • Frequency: Measured in hertz (Hz), frequency refers to the number of wave cycles that pass a point in one second.
  • Wavelength: This is the distance between two consecutive points of the same phase on the wave, like crest to crest.
  • Amplitude: Determines the volume of the sound; higher amplitude means louder sound.
Phase Difference
Phase difference describes how far along two sound waves are in their cycle relative to each other. It's crucial in understanding how waves interact through interference. When two waves meet, their combined effect depends on their phase relationship.

  • In Phase: When two waves have a phase difference of 0 or multiples of 2蟺, they're in sync, leading to constructive interference. This results in louder sound at points where they meet.
  • Out of Phase: A phase difference of 蟺 (or odd multiples of 蟺) means waves are out of sync, causing destructive interference. This can lead to reduced sound intensity or even silence.

In the loudspeakers exercise, the phase difference was calculated to determine how the speakers' sound waves interact at point C. The formula used, \(\Delta \phi = \frac{2\pi (d_B - d_A)}{\lambda}\), calculates this difference based on the speakers' distances to the point of interest and the sound wavelength.
Sound Intensity
Sound intensity refers to the sound power per unit area. It's a measure of how much sound energy passes through a certain area and is measured in watts per square meter (W/m虏). This concept helps determine how loud a sound will be at a given point.

The intensity of sound diminishes as it travels away from the source. The formula \(I = \frac{P}{4\pi r^2}\) helps calculate this by taking into account the power output of the source and the distance from the source.

In the context of the exercise, we calculated the intensity of sound reaching point C from each speaker separately. This allows us to understand how the sound's power is distributed as it travels from the loudspeakers.
Decibels
Decibels (dB) are a logarithmic unit used to measure sound intensity levels. This scale helps manage the wide range of sound intensities that humans can hear. The human ear perceives sound on a logarithmic scale, making decibels a fitting unit for sound comparison.

The formula used to calculate sound intensity levels is \(L = 10 \log_{10} \left( \frac{I}{I_0} \right)\), where \(I\) is the sound intensity and \(I_0 = 10^{-12} \, \text{W/m}^2\) is the reference intensity, the quietest sound a typical human ear can hear.

In the loudspeakers exercise, we used decibels to express how loud the sound was from each speaker and when both speakers were on. Understanding decibels is important for comprehending how changes in intensity affect our perception of loudness.

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Most popular questions from this chapter

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

A woman stands at rest in front of a large, smooth wall. She holds a vibrating tuning fork of frequency \(f_{0}\) directly in front of her (between her and the wall). (a) The woman now runs toward the wall with speed \(v_{\mathrm{W}}\) . She detects beats due to the interference between the sound waves reaching her directly from the fork and those reaching her after being reflected from the wall. How many beats per second will she detect? (Note: If the beat frequency is too large, the woman may have to use some instrumentation other than her ears to detect and count the beats.) (b) If the woman instead runs away from the wall, holding the tuning fork at her back so it is between her and the wall, how many beats per second will she detect?

A small sphere of radius \(R\) is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of \(R-\Delta R\) and a maximum of \(R+\Delta R\) with frequency \(f .\) This produces sound waves in the surrounding air of density \(\rho\) and bulk modulus \(B\) (a) Find the intensity of sound waves at the surface of the sphere. (The amplitude of oscillation of the sphere is the same as that of the air at the surface of the sphere.) (b) Find the total acoustic powcr radiated by the sphcrc. (c) At a distance \(d \gg R\) from the center of the sphere, find the amplitude, pressure amplitude, and intensity of the sound wave.

A person is playing a small fute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is 344.0 \(\mathrm{m} / \mathrm{s}\) , for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

A railroad train is traveling at 30.0 \(\mathrm{m} / \mathrm{s}\) in still air. The frequency of the note cmitted by the train whistle is 262 \(\mathrm{Hz}\) . What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?

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