91Ó°ÊÓ

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

Step by step solution

01

Calculate the distance from speaker A to point P1

Using the Pythagorean theorem, we can find the distance \(r_{A1}\) from speaker A to point \(P_1\): \(r_{A1} = \sqrt{x_1^2 + y_1^2} = \sqrt{4.50^2 + 0^2} = \sqrt{20.25} = 4.5 \mathrm{~m}\)

02

Find the intensity at P1 due to speaker A

To find the intensity \(I_{A1}\) at \(P_1\), we use the formula: \(I_{A1} = \frac{P}{4 \pi r_{A1}^2}\) where P is the power of the speaker (100 W) and \(r_{A1}\) is the distance from speaker A to point \(P_1\): \(I_{A1} = \frac{100}{4 \pi (4.5)^2} = 1.57 \times 10^{-3} \mathrm{~W/m^2}\)

03

Convert the intensity into decibels

To find the sound level \(\beta_{A1}\) in decibels at \(P_1\), we use the formula: \(\beta_{A1} = 10 \log_{10}\left(\frac{I_{A1}}{I_0}\right)\) where \(I_0\) is the reference intensity, which is \(10^{-12} \mathrm{~W/m^2}\): \(\beta_{A1} = 10 \log_{10}\left(\frac{1.57 \times 10^{-3}}{10^{-12}}\right) = 121.0 \mathrm{~dB}\)

04

Find the sound wavelength

To find the wavelength \(λ\) of the sound waves, we use the formula: \(\lambda = \frac{v}{f}\) where \(v\) is the speed of sound (approximately \(343 \mathrm{~m/s}\)) and \(f\) is the frequency of the sound waves: \(\lambda = \frac{343}{10,000} = 0.0343 \mathrm{~m}\)

05

Determine the change in distance for single minimum intensity

As the source (speakers) and receiver (point P2) positions are known, we can define the path difference between two sound waves coming from A and B as: \(\Delta L = (r_{B2} - r_{A2})\) For constructive interference at P1 and destructive interference at P2, we can equate the path difference \(\Delta L\) to an odd multiple of half-wavelength (\(\frac{\lambda}{2}\)): \(\Delta L = \left(n + \frac{1}{2}\right) \frac{\lambda}{2}\) Now, we can use the approximation formula \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\), to find an expression for \(\Delta L\): \(\Delta L = \sqrt{L^2 + D^2} - \sqrt{L^2 + (D - \Delta y)^2} \approx \frac{\Delta y D}{2\sqrt{L^2 + D^2}}\) Substitute the values for L, D, and λ into the above equation, and solve for \(\Delta y\): \(\Delta y = \frac{2 \cdot \Delta L \sqrt{L^2 + D^2}}{D}\) \(\Delta y = \frac{2 (\left(n + \frac{1}{2}\right) \frac{0.0343}{2}) \sqrt{4.5^2 + 3.6^2}}{3.6} = 0.682 (2n+1)\) We're looking for the first minimum intensity, so we let n = 0: \(\Delta y = 0.682\)

06

Calculate the y position of P2

To find the \(y\) position of \(P_2\), we simply add the \(\Delta y\) to the \(y\) position of \(P_1\): \(y_2 = y_1 + \Delta y = 0 + 0.682 = -0.682 \mathrm{~m}\) So, the values for intensity, sound level, and \(\Delta y\) are: \(I_{A1} = 1.57 \times 10^{-3} \mathrm{~W/m^2}\) \(\beta_{A1} = 121.0\, \mathrm{dB}\) \(\Delta y = 0.682 \mathrm{~m}\)

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Key Concepts

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

Intensity of Sound

The intensity of sound is a measure of how much sound energy passes through a certain area each second. It's a crucial concept for understanding how loud a sound is, as it quantifies the power of sound waves that reach your ear or any point in space. For a point source, the intensity can be calculated using the formula:

\[ I = \frac{P}{4 \pi r^2} \]
where:

• P is the power of the sound source (in watts).
• r is the distance from the source to the point (in meters).

Sound intensity decreases with distance because the energy spreads over a larger surface area. In our exercise, a 100-watt sound source creates a sound intensity at point \( P_1 \), which helps us determine how loud the sound is at that location.

Decibel Levels

Decibels (dB) are a way to express sound intensity levels. This scale is logarithmic, which means every 10-decibel increase represents a tenfold increase in intensity. The decibel level can be calculated from intensity using the formula:

\[ \beta = 10 \log_{10}\left(\frac{I}{I_0}\right)\]

• I is the sound intensity being measured.
• I₀ is the reference intensity, which is often \( 10^{-12} \mathrm{~W/m^2} \), the threshold of hearing.

In the given problem, the sound at point \( P_1 \) is expressed in decibels to provide a practical way to compare loudness levels. Understanding decibel levels is essential for applications in acoustics, as it helps us describe how loud or soft a sound is in a way that matches human perception.

Path Difference

Path difference plays a critical role in understanding sound interference, as it refers to the difference in distances traveled by two sound waves arriving at a point. This difference determines whether the waves will interfere constructively or destructively. In our exercise, the path difference affects the intensity pattern produced by two speakers.

When sound waves take different paths to reach the same point, their path difference is given by:
\[ \Delta L = r_2 - r_1 \]
where \( r_2 \) and \( r_1 \) are the distances from each sound source to the point. Depending on this path difference and the wavelength, the waves can add up to make a louder sound (constructive interference) or cancel each other out (destructive interference).

Constructive and Destructive Interference

Constructive and destructive interference are fundamental concepts in wave physics. When two waves meet, they superimpose, resulting in different interference patterns. Constructive interference occurs when two waves are in phase and their peaks and troughs align, leading to an amplified wave.

For constructive interference:

• The path difference \( \Delta L \) is an integer multiple of the wavelength \( \lambda \): \( n\lambda \).

On the other hand, destructive interference happens when waves are out of phase, specifically when their peaks align with the troughs of the other wave, causing cancellation.

For destructive interference:

• The path difference \( \Delta L \) is an odd multiple of half-wavelengths: \( (n + 0.5)\lambda \).

Understanding these interference types helps explain why sounds can get softer or louder depending on their positions relative to the sources, such as the variation observed from point \( P_1 \) to \( P_2 \) in the problem.