Resultant equation of two identical out of phase waves

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Homework Help Overview

The discussion revolves around the resultant amplitude of two identical traveling waves that are out of phase by π/5.0 radians. Participants explore the relationship between the resultant amplitude and the common amplitude of the waves.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive the resultant amplitude using trigonometric identities and expresses the result as a ratio. Some participants confirm the correctness of this approach, while others inquire about alternative methods or formatting options for presenting equations.

Discussion Status

The discussion includes confirmations of the original poster's calculations, with participants providing feedback on presentation style. There is an exploration of LaTeX as a formatting tool for equations, indicating a productive exchange of ideas regarding clarity in mathematical communication.

Contextual Notes

Participants discuss the clarity of mathematical expressions and the potential for using LaTeX for better presentation, suggesting an interest in improving communication of complex ideas.

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Homework Statement


Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resultant wave in terms of the common amplitude ym of the two combining waves? (Give the answer as the ratio of the total amplitude to the common amplitude.)

Homework Equations


Let y=ym.sinx be the equation of a wave where x is a variable time dependent. and ym is the maximum amplitude.
Let y=ym.sin(x+ π/5.0) be the equation of a second identical wave but out of phase by π/5.0
In trigonometric equations: sin(a)+sin(b)=2sin(0.5(a+b)).cos(0.5(a-b))

The Attempt at a Solution


the resultant equation will be:
Y=ymsinx + ym.sin(x+ π/5.0)
Y=ym(sinx + sin(x+ π/5.0))
....but sinx + sin(x+ π/5.0)=2sin(0.5(x+x+π/5.0)).cos(0.5(x-x-π/5.0))
....=2sin(x+π/10.0).cos(π/10.0)
Y=ym(2sin(x+π/10.0).cos(π/10.0))
then Ym=2ym.cos(π/10.0) therefore Ym/ym=2 cos(π/10.0)=1.902
is that correct ? and is that the ratio they want?
 
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Yes, this is correct. You may want to use LaTeX in the future to make your work more legible, but good work.
 
anlon said:
Yes, this is correct. You may want to use LaTeX in the future to make your work more legible, but good work.
What do you mean LateX? Is there another way to solve it?
 
##\LaTeX## is a formatting system for writing equations and papers that look nice. For example, instead of writing:
Ym=2ym.cos(π/10.0) therefore Ym/ym=2 cos(π/10.0)=1.902
You can write $$Y_m = 2y_m \cos{\left(\frac{\pi}{10}\right)} \Rightarrow \frac{Y_m}{y_m} = 2 \cos{\left(\frac{\pi}{10}\right)} = 1.902$$This can also be written inline with other text: ##Y_m = 2y_m \cos{\left(\frac{\pi}{10}\right)} \Rightarrow \frac{Y_m}{y_m} = 2 \cos{\left(\frac{\pi}{10}\right)} = 1.902## which is often useful. In the bottom left corner of the reply box there is a link to the forum LaTeX guide, which tells you how to do this in the forum (wrap standalone equations in "$$" without quotes and wrap inline equations with "##" without quotes).
 
Okay I'll try it next time. Thanks :)
 

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