Chapter 18, Problem 18.13P

Two identical loudspeakers 10.0 m apart are driven by the same oscillator with a frequency of f = 21.5 Hz (Fig. P17.6) in an area where the speed of sound is 344 m/s. (a) Show that a receiver at point A records a minimum in sound intensity from the two speakers. (b) If the receiver is moved in the plane of the speakers, show that the path it should take so that the intensity remains at a minimum is along the hyperbola 9x² − 16y² = 144 (shown in red-brown in Fig. P17.6). (c) Can the receiver remain at a minimum and move very far away from the two sources? If so, determine the limiting form of the path it must take. If not, explain how far it can go.

Figure P17.6

To determine

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

Answer to Problem 18.13P

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.

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 18.13P

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

To determine

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

Answer to Problem 18.13P

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