At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.08 cos(1.7 t - 2.5 x) between two points which have zero displacement (y = 0). Find the total force exerted by the rest of the rope on this section. Neglect any effects due to the weight of the rope. Use the small-angle approximation where q, sin(q), and tan(q) are all approximately equal to each other. Mue = .3 kg/m.
y = Acos(kx-wt)
k = 2pi/lamda
w = pi*f
v^2 = F/(mue) = (tension) / (mass per unit length)
v = w/k
-kAsin(kx-wt) ~= theta
The Attempt at a Solution
I started out by recognizing the wave equation as being almost in the form y = Acos (kx - wt). So I changed it to y = Acos(-2.5x+1.7t). This equation gives the k and w values. Then, I found the tension using v^2 = F/(mue) and v = w/k. Since I knew that the force in the x direction had to cancel out because the wave isn't moving in the x direction, I used a force in the y direction equation. For this I got Fy = 2Tsin(theta). Plugging in my tension and found angle Fy = 2*.1378N*sin(.13558) = .03737, which is not the correct answer.
I would appreciate any help I can get, thank you.