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High School Math Algebra Holt Mcdougal Larson Algebra 2: Student Edition 2012

The resultant of two waves y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 is 2 A cos 2 π t 3 cos 2 π x 5 . Given data: Equations of two wave travelling through a taught string that form a standing wave are y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 , and displacement of the standing wave from these waves is y = y 1 + y 2 or y = 2 A cos 2 π t 3 cos 2 π x 5 . Method/Formula used: Two angles sum and difference formula for cosine is given by cos A − B = cos A cos B + sin A sin B cos A + B = cos A cos B − sin A sin B ... (1) Proof: The two given waves that form a standing wave are y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 that produce a resultant displacement y at each point of the taught string given by y = y 1 + y 2 … (2) Substitute A cos 2 π t 3 − 2 π x 5 for y 1 and A cos 2 π t 3 + 2 π x 5 for y 2 in equation (2), y = A cos 2 π t 3 − 2 π x 5 + A cos 2 π t 3 + 2 π x 5 = A cos 2 π t 3 − 2 π x 5 + cos 2 π t 3 + 2 π x 5 Use formulae (1) by treating 2 π t 3 as A , and 2 π x 5 as B , y = A cos 2 π t 3 − 2 π x 5 + cos 2 π t 3 + 2 π x 5 = A cos 2 π t 3 cos 2 π x 5 + sin 2 π t 3 sin 2 π x 5 + cos 2 π t 3 cos 2 π x 5 − sin 2 π t 3 sin 2 π x 5 = A 2 cos 2 π t 3 cos 2 π x 5 = 2 A cos 2 π t 3 cos 2 π x 5 Thus, resultant displacement/amplitude is y 1 + y 2 equals y = 2 A cos 2 π t 3 cos 2 π x 5 . Hence, the result.

The resultant of two waves y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 is 2 A cos 2 π t 3 cos 2 π x 5 . Given data: Equations of two wave travelling through a taught string that form a standing wave are y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 , and displacement of the standing wave from these waves is y = y 1 + y 2 or y = 2 A cos 2 π t 3 cos 2 π x 5 . Method/Formula used: Two angles sum and difference formula for cosine is given by cos A − B = cos A cos B + sin A sin B cos A + B = cos A cos B − sin A sin B ... (1) Proof: The two given waves that form a standing wave are y 1 = A cos 2 π t 3 − 2 π x 5 and y 2 = A cos 2 π t 3 + 2 π x 5 that produce a resultant displacement y at each point of the taught string given by y = y 1 + y 2 … (2) Substitute A cos 2 π t 3 − 2 π x 5 for y 1 and A cos 2 π t 3 + 2 π x 5 for y 2 in equation (2), y = A cos 2 π t 3 − 2 π x 5 + A cos 2 π t 3 + 2 π x 5 = A cos 2 π t 3 − 2 π x 5 + cos 2 π t 3 + 2 π x 5 Use formulae (1) by treating 2 π t 3 as A , and 2 π x 5 as B , y = A cos 2 π t 3 − 2 π x 5 + cos 2 π t 3 + 2 π x 5 = A cos 2 π t 3 cos 2 π x 5 + sin 2 π t 3 sin 2 π x 5 + cos 2 π t 3 cos 2 π x 5 − sin 2 π t 3 sin 2 π x 5 = A 2 cos 2 π t 3 cos 2 π x 5 = 2 A cos 2 π t 3 cos 2 π x 5 Thus, resultant displacement/amplitude is y 1 + y 2 equals y = 2 A cos 2 π t 3 cos 2 π x 5 . Hence, the result.

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Holt Mcdougal Larson Algebra 2: Student Edition 2012

1st Edition

ISBN: 9780547647159

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Expert Solution & Answer

Chapter 10.6, Problem 42PS

Solution

To show: The resultant of two waves y1=Acos2πt3−2πx5 and y2=Acos2πt3+2πx5 is 2Acos2πt3cos2πx5 .

Given data: Equations of two wave travelling through a taught string that form a standing wave are y1=Acos2πt3−2πx5 and y2=Acos2πt3+2πx5 , and displacement of the standing wave from these waves is y=y1+y2 or y=2Acos2πt3cos2πx5 .

Method/Formula used:

Two angles sum and difference formula for cosine is given by

cosA−B=cosAcosB+sinAsinBcosA+B=cosAcosB−sinAsinB ... (1)

Proof:

The two given waves that form a standing wave are y1=Acos2πt3−2πx5 and y2=Acos2πt3+2πx5 that produce a resultant displacement y at each point of the taught string given by

y=y1+y2 … (2)

Substitute Acos2πt3−2πx5 for y1 and Acos2πt3+2πx5 for y2 in equation (2),

y=Acos2πt3−2πx5+Acos2πt3+2πx5=Acos2πt3−2πx5+cos2πt3+2πx5

Use formulae (1) by treating 2πt3 as A , and 2πx5 as B ,

y=Acos2πt3−2πx5+cos2πt3+2πx5=Acos2πt3cos2πx5+sin2πt3sin2πx5+cos2πt3cos2πx5−sin2πt3sin2πx5=A2cos2πt3cos2πx5=2Acos2πt3cos2πx5

Thus, resultant displacement/amplitude is y1+y2 equals y=2Acos2πt3cos2πx5 .

Hence, the result.

Chapter 10.6, Problem 41PS

Chapter 10.6, Problem 43PS

Chapter 10 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 10.1 - Prob. 1GP Ch. 10.1 - Prob. 2GP Ch. 10.1 - Prob. 3GP Ch. 10.1 - Prob. 4GP Ch. 10.1 - Prob. 5GP Ch. 10.1 - Prob. 6GP Ch. 10.1 - Prob. 7GP Ch. 10.1 - Prob. 8GP Ch. 10.1 - Prob. 9GP Ch. 10.1 - Prob. 10GP

Ch. 10.1 - Prob. 11GP Ch. 10.1 - Prob. 12GP Ch. 10.1 - Prob. 13GP Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29PS Ch. 10.1 - Prob. 30PS Ch. 10.1 - Prob. 31PS Ch. 10.1 - Prob. 32PS Ch. 10.1 - Prob. 33PS Ch. 10.2 - Prob. 1GP Ch. 10.2 - Prob. 2GP Ch. 10.2 - Prob. 3GP Ch. 10.2 - Prob. 4GP Ch. 10.2 - Prob. 5GP Ch. 10.2 - Prob. 6GP Ch. 10.2 - Prob. 7GP Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - 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