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 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 $344\text{m}/\text{s}$, frequency generated by the loudspeakers is $21.5\text{Hz}$ and the distance between two speakers is $10.0\text{m}$.

The given figure of two loudspeakers is shown below.

Figure (I)

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

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

$10\text{m}-\text{9}\text{m}=\text{1}\text{m}$

Formula to calculate the path difference for minimum intensity is,

$d_1-d_2=\frac{\lambda }{2}$ (I)

• $d_1$ is the distance from the first speaker to $A$.
• $d_2$ is the distance from the second speaker to $B$.

Substitute $\frac{v}{f}$ for $\lambda$ in equation (I).

$d_1-d_2=\frac{\frac{v}{f}}{2}$ (II)

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

Substitute $344\text{m}/\text{s}$ for $v$, $21.5\text{Hz}$ for $f$, $1\text{m}$ for $d_2$ and $9\text{m}$ for $d_1$ in equation (II).

$\begin{array}{c}9\text{m}-1\text{m}=\frac{\frac{344\text{m}/\text{s}}{21.5\text{Hz}}}{2}\\ 8\text{m=8}\text{m}\end{array}$

The difference between distance $d_1$ and $d_2$ 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.

Answer to Problem 12P

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x^2-16y^2=144$.

Explanation of Solution

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

Formula to calculate the wavelength of each wave is,

$\lambda =\frac{v}{f}$

Substitute $344\text{m}/\text{s}$ for $v$ and $21.5\text{Hz}$ for $f$.

$\begin{array}{c}\lambda =\frac{344\text{m}/\text{s}}{21.5\text{Hz}}\\ =16\text{m}\end{array}$

Thus, the wavelength of the each wave is $16\text{m}$.

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

Formula to calculate the path difference for minimum intensity is,

$d_1-d_2=\frac{\lambda }{2}$ (III)

The distance for left side speaker at point $\left(-5,0\right)$ is,

$d_1=\sqrt{{\left(x+5\right)}^2+y^2}$

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

$d_2=\sqrt{{\left(x-5\right)}^2+y^2}$

Substitute $\sqrt{{\left(x+5\right)}^2+y^2}$ for $d_1$ and $\sqrt{{\left(x-5\right)}^2+y^2}$ for $d_2$ in equation (III).

$\begin{array}{c}\sqrt{{\left(x+5\right)}^2+y^2}-\sqrt{{\left(x+5\right)}^2+y^2}=\frac{\lambda }{2}\\ \sqrt{{\left(x+5\right)}^2+y^2}=\frac{\lambda }{2}+\sqrt{{\left(x+5\right)}^2+y^2}\end{array}$ (IV)

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

$\begin{array}{c}{\left(\sqrt{{\left(x+5\right)}^2+y^2}\right)}^2={\left(\frac{\lambda }{2}+\sqrt{{\left(x+5\right)}^2+y^2}\right)}^2\\ 20x-\frac{{\lambda }^2}{4}=\lambda \sqrt{{\left(x-5\right)}^2+y^2}\end{array}$ (V)

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

$\begin{array}{c}{\left(20x-\frac{{\lambda }^2}{4}\right)}^2={\left(\lambda \sqrt{{\left(x-5\right)}^2+y^2}\right)}^2\\ 400x^2+\frac{{\lambda }^4}{16}-10x\lambda ={\lambda }^2\left({\left(x-5\right)}^2+y^2\right)\end{array}$ (VI)

Substitute $16\text{m}$ for $\lambda$ in equation (VI).

$\begin{array}{c}400x^2+\frac{{\left(16\text{m}\right)}^4}{16}-10x\times 16\text{m}={\left(16\text{m}\right)}^2\left({\left(x-5\right)}^2+y^2\right)\\ 9x^2-16y^2=144\end{array}$

Conclusion:

Therefore, the path takes by the receiver is along the hyperbola whose equation is given as, $9x^2-16y^2=144$.

(c)

To determine

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

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 $\pm \frac{3}{4}$.

Explanation of Solution

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

The equation of the hyperbola is given as,

$9x^2-16y^2=144$

Rearrange the above equation for $y$,

$y=\pm \frac{3}{4}x\sqrt{1-\frac{16}{x^2}}$

For very large value of $x$,

$y=\pm \frac{3}{4}x$

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 $\pm \frac{3}{4}$.

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 $\pm \frac{3}{4}$.