The receiver at point A records a minimum in sound intensity from the two speakers.

Question

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Answer 1

Question

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

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Students have asked these similar questions

Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?

* Coulomb's Law Example Three charges are positioned as seen below. Charge 1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is - 6.0MC. What is the magnitude and the direction of the force on charge 2 due to charges 1 and 3? 93 kq92 F == 2 r13 = 0.090m 91 r12 = 0.12m 92 Coulomb's Constant: k = 8.99x10+9 Nm²/C² ✓

Make sure to draw a Free Body Diagram as well

Answer 2

Question

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

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Answer 3

Question

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

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Answer 4

Question

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

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Answer 5

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Answer 6

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Answer 7

Chapter 14, Problem 12P

(a)

To determine

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

Answer 8

(a)

Expert Solution

Answer to Problem 12P

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10.0 m$ .

The given figure of two loudspeakers is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 14, Problem 12P

Figure (I)

From Figure (I), the distance from the first speaker to $A$ is $9 m$ .

The distance from the second speaker to $B$ is given as,

$10 m−9 m=1 m$

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (I)

$d1$ is the distance from the first speaker to $A$ .
$d2$ is the distance from the second speaker to $B$ .

Substitute $vf$ for $λ$ in equation (I).

$d1−d2=vf2$ (II)

$λ$ is the wavelength of each wave.
$v$ is the velocity of the sound.
$f$ is the frequency generated by the loudspeakers.

Substitute $344 m/s$ for $v$ , $21.5 Hz$ for $f$ , $1 m$ for $d2$ and $9 m$ for $d1$ in equation (II).

$9 m−1 m=344 m/s21.5 Hz28 m=8 m$

The difference between distance $d1$ and $d2$ of loudspeaker from point $A$ is same as the half of the wavelength. Hence, this is condition for the destructive interference. So, the receiver at point $A$ records a minimum intensity.

Conclusion:

Therefore, the receiver at point $A$ records the minimum sound intensity because of destructive interference.

Answer 13

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Answer 14

(b)

To determine

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

Answer 15

(b)

Expert Solution

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is , the distance between two speakers is $10 m$ .

Formula to calculate the wavelength of each wave is,

$λ=vf$

Substitute $344 m/s$ for $v$ and $21.5 Hz$ for $f$ .

$λ=344 m/s21.5 Hz=16 m$

Thus, the wavelength of the each wave is $16 m$ .

From figure (I), the coordinates for left side speaker is $(−5,0)$ and for the right side speaker is $(5,0)$ .

Formula to calculate the path difference for minimum intensity is,

$d1−d2=λ2$ (III)

The distance for left side speaker at point $(−5,0)$ is,

$d1=(x+5)2+y2$

The distance for the right side speaker at point $(5,0)$ is,

$d2=(x−5)2+y2$

Substitute $(x+5)2+y2$ for $d1$ and $(x−5)2+y2$ for $d2$ in equation (III).

$(x+5)2+y2−(x+5)2+y2=λ2(x+5)2+y2=λ2+(x+5)2+y2$ (IV)

Square on the both sides of the equation (IV),

$((x+5)2+y2)2=(λ2+(x+5)2+y2)220x−λ24=λ(x−5)2+y2$ (V)

Square on the both sides of the equation (V),

$(20x−λ24)2=(λ(x−5)2+y2)2400x2+λ416−10xλ=λ2((x−5)2+y2)$ (VI)

Substitute $16 m$ for $λ$ in equation (VI).

$400x2+(16 m)416−10x×16 m=(16 m)2((x−5)2+y2)9x2−16y2=144$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x2−16y2=144$ .

Answer 20

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Answer 21

(c)

To determine

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

Answer 22

(c)

Expert Solution

Answer to Problem 12P

Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information: The speed of the sound is $344 m/s$ , frequency generated by the loudspeakers is $21.5 Hz$ and the distance between two speakers is $10 m$ .

The equation of the hyperbola is given as,

$9x2−16y2=144$

Rearrange the above equation for $y$ ,

$y=±34x1−16x2$

For very large value of $x$ ,

$y=±34x$

It means the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Conclusion:Therefore, the receiver remains at a minimum and move very far away from the two sources and the limiting form of the path is two straight lines through the origin with slope $±34$ .

Answer 27

Ch. 14.1 - Prob. 14.1QQ Ch. 14.2 - Prob. 14.2QQ Ch. 14.3 - When a standing wave is set up on a string fixed...Ch. 14.4 - Prob. 14.4QQ Ch. 14.4 - Prob. 14.5QQ Ch. 14.5 - You are tuning a guitar by comparing the sound of...Ch. 14 - A flute has a length of 58.0 cm. If the speed of...Ch. 14 - Prob. 2OQ Ch. 14 - In Figure OQ14.3, a sound wave of wavelength 0.8 m...Ch. 14 - Prob. 4OQ

The receiver at point A records a minimum in sound intensity from the two speakers.

Concept explainers

Videos

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.

Answer to Problem 12P

Explanation of Solution

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

Answer to Problem 12P

Explanation of Solution

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

Answer to Problem 12P

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

The receiver at point A records a minimum in sound intensity from the two speakers.

Concept explainers

Videos

To show: The receiver at point A<m:math><m:mi>A</m:mi> </m:math> records a minimum in sound intensity from the two speakers.

Answer to Problem 12P

Explanation of Solution

To show: The path it should take so that the intensity remains at a minimum is along the given hyperbola equation.

Answer to Problem 12P

Explanation of Solution

To explain:Whether the receiver remains at a minimum and move very far away from the two sources.

Answer to Problem 12P

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

To show: The receiver at point $A$ records a minimum in sound intensity from the two speakers.