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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