Finding Two Wave's Common Amplitude

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SUMMARY

The discussion centers on the analysis of two sinusoidal waves traveling along a string, represented by the equation y'(x, t) = (1.0 mm) sin(18x - 4.0t + 0.960 rad). The wavelength λ was determined to be 0.35 meters, and the phase difference was calculated as 1.92 radians. The amplitude ym was initially miscalculated but was correctly found to be approximately 0.872 mm by equating the given wave equation to the sum equation and solving for ym using the sine function.

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Homework Statement


Two sinusoidal waves, identical except for phase, travel in the same direction along a string producing a net wave y'(x, t) = (1.0 mm) sin(18x - 4.0t + 0.960 rad), with x in meters and t in seconds.

(a) What is the wavelength λ of the two waves?
(b) What is the phase difference between them?
(c) What is their amplitude ym?

I got A and B correct with answers of 0.35 and 1.92

Homework Equations


y'm = 2ym cos([tex]\frac{\Phi}{2}[/tex])

The Attempt at a Solution



I set y'm = 1 mm and [tex]\frac{\Phi}{2}[/tex] = 0.96

Then

[tex]\frac{1}{2cos(.96)}[/tex] = ym

0.5 = ym (Rounding)

Not sure what I'm doing wrong.

Any help is appreciated.

Thanks!
 
Last edited:
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I found my answer. I took the given equation and set it equal to the sum equation.

I then set x and t to zero and solved for ym

That led to

[tex]\frac{sin(\frac{\Phi}{2})}{sin(\Phi)}[/tex]

In this case


[tex]\frac{sin(.96)}{sin(1.92)}[/tex] = ym = .872
 

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