Solving the Transverse Wave on a Rope: Tension and Force

[masamune]

The transverse displacement of an harmonic wave on a stretched rope is y = 0.04 cos(2.5 t - 3.3 x), where x and y are in meters and t is in seconds. A 5 meter length of this rope has a mass of 1.5 kg.

a) What is the tension in the rope?
b) At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.04 cos(2.5 t - 3.3 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.(see attached picture)

I managed to get part a by using the general equation y = Acos(kx-wt) where A is amplitude, k is the "wave number", and w (omega) is the angular frequency. Therefore, I was able to use v=sqrt(T/u). u is the mass per unit length of the rope (1.5/5) Then I found the tension of the rope to be 0.1716

For part b, I realize that the tension due to the left half of the rope pulling the loop is down and to the left and the tension due to the right half of the rope pulls the loop down and to the right. I can't figure out what is the net force on the loop or what direction it points. Thanks for any help you can offer.

 

[HallsofIvy]

You have already calculated the tension,T, on the rope. Take it to be horizontal and find the vertical component of force from tan(θ)= F/T. The horizontal components of force, being T left and right, will cancel and the net force is 2F.

 

[masamune]

How do I find the angle? What does theta correspond to in the first place? I understand that the left and right horizontal components of the force cancel though. I tried substituting 90 degrees since the two forces are perpendicular to each other, however this didn't give me a correct answer..could you please elaborate?

 

[HallsofIvy]

You are told that the "at t=0, two points have 0 displacement"- that is, y= 0.04 cos(2.5 t - 3.3 x) = 0.04 cos(0-3.3x)= 0.04 cos(3.3x)= 0 or cos(3.3x)= 0 so you can find x. (3.3 x= -π/2 and 3.3 x= π/2.). The tangent of the angle (i.e. the slope) is the derivative with respect to x there: y'= (0.04)(-sin(2.5t- 3.3x)(-3.3) which, at t=0 and x= +/- π/6.6, is: 0.132 sin(+/- π/2)= -0.132 and +0.132. You don't actually need the angle, just tan(θ)= 0.132. The force on the right is given by
F/T= 0.132 and the net tension is 2F.