Finding Two Wave's Common Amplitude

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The discussion revolves around finding the common amplitude of two sinusoidal waves traveling along a string. The net wave is given by the equation y'(x, t) = (1.0 mm) sin(18x - 4.0t + 0.960 rad). The user correctly calculated the wavelength as 0.35 m and the phase difference as 1.92 rad. Initially, there was confusion regarding the amplitude calculation, but the user eventually derived the correct amplitude using the relationship ym = sin(Φ/2) / sin(Φ), resulting in ym = 0.872. The problem illustrates the process of resolving wave properties through trigonometric relationships.
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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(\frac{\Phi}{2})

The Attempt at a Solution



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

Then

\frac{1}{2cos(.96)} = ym

0.5 = ym (Rounding)

Not sure what I'm doing wrong.

Any help is appreciated.

Thanks!
 
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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

\frac{sin(\frac{\Phi}{2})}{sin(\Phi)}

In this case


\frac{sin(.96)}{sin(1.92)} = ym = .872
 
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