What is the issue with the wave equation for a flexible cable including gravity?

[daudaudaudau]

Hi.

I think the wave equation for a flexible cable including gravity should look like this
[tex]
\frac{\partial^2}{\partial x^2}f(x,t)-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}f(x,t)=g
[/tex]
It this true? (g is the gravitational constant)

Now if I put the boundary conditions [itex]f(x=0,t)=0 [/itex], [itex]f(x=1,t)=0[/itex] and [itex]f(x,t=0)=0[/itex] a solution to the equation would be [itex]f(x,t)=\frac{g}{2}x(x-1)[/itex]. But this tells me that the cable will follow a parabola under the influence of gravity, which is not true. What is the problem?

 

[hikaru1221]

I guess you're finding the shape of the stable cable, which accounts for your solution at which t is not present. If so, then yes, the shape is a parabola. Why is it so obviously wrong?
By the way, I think the wave equation is wrong in dimension, which leads to that the solution is wrong in dimension too.

 

[Cleonis]

I guess you're finding the shape of the stable cable, which accounts for your solution at which t is not present. If so, then yes, the shape is a parabola.

Actually, the shape of a cable suspended at both ends is a catenary, not a parabola. Solving for the catenary gives the hyperbolic cosine.

[tex] cosh(x) = \tfrac{1}{2}(e^x + e^{-x}) [/tex]

 

[Vanadium 50]

First, it's a catenary.

Second, why did you start with a 1-dimensional wave equation in x and t and expect it to give you y vs. x?

 

[hikaru1221]

Actually, the shape of a cable suspended at both ends is a catenary, not a parabola. Solving for the catenary gives the hyperbolic cosine.

[tex] cosh(x) = \tfrac{1}{2}(e^x + e^{-x}) [/tex]

I think it's [tex]y=\frac{1}{a} cosh(ax)=\frac{1}{2a}(e^{ax}+e^{-ax})[/tex].
For cables which are hung tightly, a is small and therefore, we end up with that y(x) is a parabola approximately. Though I don't think this is valid because the assumption we have to make at the start in order to arrive at catenary is that the cable is hung gently, my intuition tells me that the actual result if the cable is not hung gently is parabola.Now the wave equation derived by the OP is: [tex]
\frac{\partial^2}{\partial x^2}f(x,t)-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}f(x,t)=g
[/tex]
I couldn't derive this equation without one condition: the cable is not hung gently, i.e. [tex]tan\theta = \frac{\partial f}{\partial x} << 1[/tex]. I guess the "flexibility" condition means that the cable is not too hard and there is no sudden change in dy/dx - the curve is a smooth curve.
Under this condition, after doing some simple math, we can arrive at that y(x) is a parabola.Anyway it's just my speculation. Correct me if I'm wrong.

 

[daudaudaudau]

Second, why did you start with a 1-dimensional wave equation in x and t and expect it to give you y vs. x?

You can derive the wave equation for a flexible cable by considering a series of mass points connected by springs. So [itex]f(x,t)[/itex] is the y-coordinate of the flexible cable at position x at time t. Now in my derivation of the wave equation I tried adding in an additional force which is gravity.

 

[hikaru1221]

You can derive the wave equation for a flexible cable by considering a series of mass points connected by springs. So [itex]f(x,t)[/itex] is the y-coordinate of the flexible cable at position x at time t. Now in my derivation of the wave equation I tried adding in an additional force which is gravity.

There is a difference between the suspended cable and the series of masses and springs: while the series is in horizontal position, the cable doesn't (so you have to deal with the angle). There are 2 extreme cases of the main factor affecting the vibration of the cable: it is either gravity or tension (but not both). If it's gravity, I think we cannot write f(x,t), because the propagating direction of the wave is not the x direction. If it's tension, we can.

 

[Vanadium 50]

My point is that you want y vs. x. But you wrote down an equation with x, t and no y. See the problem?