Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 17-46, sound of wavelength $0 .850 m$ is emitted isotropically by point source S. Sound ray 1 extends directly to detector D, at distance $L = 10.0 m$ . Sound ray 2 extends to D via a reflection (effectively, a “bouncing”) of the sound at a flat surface. That reflection occurs on a perpendicular bisector to the SD line, at distance d from the line. Assume that the reflection shifts the sound wave by $0 .500 λ$ . For what least value of d (other than zero) do the direct sound and the reflected sound arrive at D (a) exactly out of phase and (b) exactly in phase?

The least value of d for which the direct and reflected sounds arrive at D exactly out of phase is, $2.10 m$ .
The least value of d for which the direct and reflected sounds arrive at D exactly in phase is $1.47 m$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Wavelength of sound emitted isotopically by point source is $0.850 m$ .
Sound ray 1 distance to detector D, $L = 10.0 m$ .
Reflection shifts the sound wave by $0.500 λ$

Step 2: Understanding the concept of interference

We can find the path difference between the direct and reflected waves. Then using the conditions for constructive and destructive interference, we can find the least value of d, for which the direct and reflected sounds arrive at D exactly in phase and out of phase.

Formulae:

The cosine law for side c of triangle,

$c^{2} = a^{2} + b^{2} - 2 abcosC$ …(i)

The linear expansion formula,

$L = L_{0} (1 + αΔT)$ …(ii)

Step 3: a) Calculation of least value of d for destructive interference

Path difference between direct and reflected wave using equations (i) and (ii) is given as:

$\begin{matrix} Δx = \sqrt{L^{2} + {(2 d)}^{2}} - L + 0.500 λ \\ = \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 10 m + 0.500 (0.850 m) \\ = \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 9.575 m \end{matrix}$

For destructive interference, the least value of d is given as:

$\begin{matrix} \frac{Δx}{λ} = 0.5, 1.5, \dots . \\ \frac{\sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 9.575 m}{0.850 m} = 0.5, 1.5, \dots . \\ \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 9.575 m = 0.425 m, 1.275 m, \dots . \\ \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} = 10 m, 10.85 m, .. \\ {(10 m)}^{2} + {(2 d)}^{2} = 100 m, 117.72 m, ... \\ d = 0, 2.1 m ... \end{matrix}$

Hence the value of d is, $0, 2.1 m ...$ .

Excluding zero, the least value is found to be $d = 2.10 m$ .

Therefore, the least value of d for which the direct and reflected sounds arrive at D exactly out of phase is $2.10 m$ .

Step 4: b) Calculation of least value of d for constructive interference

For constructive interference, the least value of d is given as:

$\begin{matrix} \frac{Δx}{λ} = 1, 2, \dots . \\ \frac{\sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 9.575 m}{0.850 m} = 1, 2, \dots . \\ \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} - 9.575 m = 0.850 m, 1.7 m, \dots . \\ \sqrt{{(10 m)}^{2} + {(2 d)}^{2}} = 10.425 m, 11.275 m, ... \\ {(10 m)}^{2} + {(2 d)}^{2} = 108.68 m, 127.126 m, .. \\ d = 1.47 m, 2.6 m, ... \end{matrix}$

Solving this, we get the least value of d as:

$d = 1.47 m$

Therefore, the least value of d for which the direct and reflected sounds arrive at D exactly in phase is $1.47 m$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of interference

Step 3: a) Calculation of least value of d for destructive interference

Step 4: b) Calculation of least value of d for constructive interference

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