Finding Two Wave's Common Amplitude

[foundAnAnswer]

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 y_m?

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

Homework Equations

y'_m = 2y_m cos($$\frac{\Phi}{2}$$)

The Attempt at a Solution

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

Then

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

0.5 = y_m (Rounding)

Not sure what I'm doing wrong.

Any help is appreciated.

Thanks!

 

[foundAnAnswer]

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 y_m

That led to

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

In this case


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

 

[kerimek]

Your approach is correct, but there is a small error in your calculation. When dividing by cos(0.96), you should use the inverse cosine function (cos^-1) rather than just 0.96. So the correct calculation would be:

ym = 1 mm / (2cos^-1(0.96)) = 0.50 mm

So the amplitude of both waves is 0.50 mm.

Hope this helps!