Q9.3P Question: Use Eq. 9.19 to determ... [FREE SOLUTION]

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Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 9: Q9.3P (page 388)

Question: Use Eq. 9.19 to determine $A_{3}$ and $未_{3}$ in terms ofrole="math" localid="1653473428327" $A_{1}$ , $A_{2}$ , $未_{1}$ , and $未_{2}$ .

Short Answer

Expert verified

The value of $A_{3}$ is role="math" localid="1653473609796" $A_{3} = \sqrt{A_{1}^{2}} + A_{2}^{2} + 2 A_{1} A_{2} \cos (未_{1} - 未_{2})$ , and the value of $未_{3}$ is $未_{3} = \tan^{- 1} [\frac{A_{1} \sin 未_{1} + A_{2} \sin 未_{2}}{A_{1} \cos 未_{1} + A_{2} \cos 未_{2}}]$

Step by step solution

Expression for the amplitude equation 9.19:

Write the amplitude equation 9.19.

$A_{3} e^{i 未_{3}} = A_{1} e^{i 未_{1}} + A_{2} e^{i 未_{2}}$

It is known that $e^{i 胃} = \cos 胃 + i \sin 胃$ . So, the equation (1) becomes,

$A_{3} (c o s 未_{3} + i s i n 未_{3}) = A_{1} (c o s 未_{1} + i s i n 未_{1}) + A_{2} (c o s 未_{2} + i s i n 未_{2})$

Determine the value of :

Solve for the real term.

$A_{3} \cos 未_{3} = A_{1} \cos 未_{1} + A_{2} \cos 未_{2}$ 鈥�.. (2)

Solve for the imaginary term.

$A_{3} \sin 未_{3} = A_{1} \sin 未_{1} + A_{2} \sin 未_{2}$ 鈥�.. (3)

Squaring equations (2) and (3) and add them.

$A_{3}^{2} \cos^{2} 未_{3} + A_{3}^{2} \sin^{2} 未_{3} = {(A_{1} \cos 未_{1} + A_{2} \cos 未_{2})}^{2} + {(A_{1} \sin 未_{1} + A_{2} \sin 未_{2})}^{2} A_{3}^{2} (\cos^{2} 未_{3} + \sin^{2} 未_{3}) = {(A_{1} \cos 未_{1} + A_{2} \cos 未_{2})}^{2} + {(A_{1} \sin 未_{1} + A_{2} \sin 未_{2})}^{2} A_{3}^{2} = {(A_{1} \cos 未_{1} + A_{2} \cos 未_{2})}^{2} + {(A_{1} \sin 未_{1} + A_{2} \sin 未_{2})}^{2} A_{3}^{2} = [(A_{1}^{2} \cos^{2} 未_{1} + A_{2}^{2} \cos^{2} 未_{2} + 2 A_{1} A_{2} \cos 未_{1} \cos 未_{2}) + A_{1}^{2} \sin^{2} 未_{1} + A_{2}^{2} \sin^{2} 未_{2} + 2 A_{1} A_{2} \sin 未_{1} \sin 未_{2}]$

Simplify further.

$A_{3}^{2} = A_{1}^{2} (\cos^{2} 未_{1} + \sin^{2} 未_{1}) + A_{2}^{2} (\cos^{2} 未_{2} + \sin^{2} 未_{2}) + 2 A_{1} A_{2} (\cos 未_{1} \cos 未_{2} + \sin 未_{1} \sin 未_{2}) A_{3}^{2} = A_{1}^{2} (1) + A_{2}^{2} (1) + 2 A_{1} A_{2} \cos (未_{1} - 未_{2}) A_{3}^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (未_{1} - 未_{2}) A_{3} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (未_{1} - 未_{2})}$

Therefore, the value of $A_{3}$ is $A_{3} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (未_{1} - 未_{2})}$ .

Determine the value of δ3 :

Divide equation (3) by (2).

$\frac{A_{3} \sin 未_{3}}{A_{3} \cos 未_{3}} = \frac{A_{1} \sin 未_{1} + A_{1} \sin 未_{2}}{A_{1} \cos 未_{1} + A_{1} \cos 未_{2}} \tan 未_{3} = \frac{A_{1} \sin 未_{1} + A_{1} \sin 未_{2}}{A_{1} \cos 未_{1} + A_{1} \cos 未_{2}} 未_{3} = \tan^{- 1} (\frac{A_{1} \sin 未_{1} + A_{1} \sin 未_{2}}{A_{1} \cos 未_{1} + A_{1} \cos 未_{2}})$

Therefore, the value of $未_{3}$ is $未_{3} = \tan^{- 1} (\frac{A_{1} \sin 未_{1} + A_{1} \sin 未_{2}}{A_{1} \cos 未_{1} + A_{1} \cos 未_{2}})$ .

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Question: Use Eq. 9.19 to determine $A_{3}$ and $未_{3}$ in terms ofrole="math" localid="1653473428327" $A_{1}$ , $A_{2}$ , $未_{1}$ , and $未_{2}$ .

Short Answer

Step by step solution

Expression for the amplitude equation 9.19:

Determine the value of :

Determine the value of δ3 :

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91影视

Question: Use Eq. 9.19 to determineA3and未3in terms ofrole="math" localid="1653473428327" A1,A2,未1, and未2.

Short Answer

Step by step solution

Expression for the amplitude equation 9.19:

Determine the value of :

Determine the value of δ3 :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Space Physics

Particle Model of Matter

Fundamentals of Physics

Further Mechanics and Thermal Physics

Physical Quantities And Units

Study anywhere. Anytime. Across all devices.

Question: Use Eq. 9.19 to determine $A_{3}$ and $未_{3}$ in terms ofrole="math" localid="1653473428327" $A_{1}$ , $A_{2}$ , $未_{1}$ , and $未_{2}$ .