What is the Total Force Exerted on a Section of a Wave-Carrying Rope?

[Ridonkulus]

Homework Statement

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.

Homework Equations

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.

 

[OlderDan]

I don't see where you are getting your sine factor. What is the slope of the rope at a point of zero displacement?

 

[andrevdh]

What I would suggest is to determine the acceleration of the rope $$\ddot{y}$$ at two x distances a half wavelength apart (both will be functions of time of cause) . Multiply this with the mass of the half wavelength piece of rope and add the two tensions that it experiences at its ends. What I am therefore suggesting amounts to consider the piece of rope responding as a unit to the two forces at its ends (in reality it deforms under these forces, but N2 is still applicable).

 

[Ridonkulus]

I'm sorry, I missed that part of my work. I used the equation -kAsin(kx-wt) (this came out to be -(-1.7)*.08*sin(-2.5*.25*2pi/-2.5)) to get the sine factor. The number I got was .136, using half of the wave length. This is what I plugged into the sin(theta).

 

[andrevdh]

The equation

$$c = \sqrt{\frac{T}{\mu}}$$

gives the tension in the rope in the x-direction (it controls the propagation of the wave along the rope).

For the SHM to exist the tension in the rope needs to oscillate with time as the disturbance propagates throught it (the y-component of the tension in the rope). In this case $$c$$ is the (constant) speed of propagation of the disturbance not the SHM speed (which varies with time).

Come to think about it one need to add the two tension components.

 

[OlderDan]

I'm sorry, I missed that part of my work. I used the equation -kAsin(kx-wt) (this came out to be -(-1.7)*.08*sin(-2.5*.25*2pi/-2.5)) to get the sine factor. The number I got was .136, using half of the wave length. This is what I plugged into the sin(theta).

That does not look right.

y(x,t) = Acos(kx-ωt) = Acos(ωt-kx) = 0.08 cos(1.7 t - 2.5 x)

I don't see that you gain anything by negating the argument of the cosine, but it's OK. You did not use the negated form to get -kAsin(kx-ωt)

You are interested in the shape of the rope at t = 0 (this time specification is not really needed in the problem; all you need is to look at adjacent points of zero displacement at some unspecified time). In particular, you need the angle between the rope and the line y = 0 at the points of intersection of y(x,0) and y = 0. Take a snapshot of the rope at t = 0 and you get

y(x,0) = 0.08 cos(2.5 x)

What you need is the slope of this function at points where y(x,0) = 0. Your expression

-kAsin(kx-ωt)

is the derivative of y(x,t) wrt x, which is the slope of the function at position x at any time t. The way the problem is stated, you want it at t = 0. So you are looking for the value of

dy(x,0)/dx = -kAsin(kx)

at points where y(x,0) = 0. Think about what has to be true if y(x,0) = 0, and use that in your slope expression.

 

[Ridonkulus]

Sorry I haven't been able to try this until just now. Thanks for all of the help everyone. I got the problem (for anyone that is curious the answer turned out to be .055128 N) using the help OlderDan gave me, thanks again.