Wave propagation along a rubber cord.

  • #1

Homework Statement



You have a rubber cord of relaxed length x. It be-
haves according to Hooke's law with a "spring con-
stant" equal to k. You then stretch the cord so it has
a new length equal to 2x. a) Show that a wave will
propagate along the cord with speed

v=[tex]\sqrt{\frac{2kx^{2}}{m}}[/tex]

b) You then stretch the cord further so that the cord's
length increases with speed v/3. Show that the wave
will propagate during the stretching with a speed that
is not constant:

v(t)=[tex]\sqrt{\frac{kx^{2}}{m}(1+t\sqrt{\frac{2k}{9m}})(2+t\sqrt{\frac{2k}{9m}})}[/tex]

Homework Equations



strings wave propagation speed: v=[tex]\sqrt{\frac{T}{u}}[/tex]

hookes law: F=-kx

Where T is tension and u is linear mass density

The Attempt at a Solution



On part A i used hookes law to obtain the tension: T = k2x (not sure how to explain the negative sign). And u=m/x ( i dont understand why would you use the orignal length of x to obtain the linear mass density instead of the new length of 2x). Basically plug that in into the equation waves propagation speed and you get the awnser.

My train of thought for part b is that if your length is changing at a constant rate of v/3 then so is thetension. The new tension would be given by
T(t)=vtk/3=([tex]\frac{tk}{3}[/tex])[tex]\sqrt{\frac{2kx^{2}}{m}}[/tex] then i plugged that in into the waves propgation speed equation for T and keeping u=m/x. And then iam stuck...iam not sure how to procceed from there any help/hint is appreciated.

Thanks :)!
 
Last edited:

Answers and Replies

  • #2
shameless bumb :( still cant solve it.
 
  • #3
Redbelly98
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The Attempt at a Solution



On part A i used hookes law to obtain the tension: T = k2x (not sure how to explain the negative sign).
Don't worry about the negative sign, that is because Hooke's Law gives the force exerted by the spring/cord on an attached object. The tension in the cord itself is not negative.

The big problem here is that, in Hooke's Law, "x" is the amount by which the cord has stretched, it is not the total length of the cord. Unfortunately, they are also using x to mean the cord's relaxed length, which is different. So we have to distinguish these two:

Let Δx = the amount by which the cord has stretched, so that
F = k Δx​
And let just plain x = the relaxed length of the cord, so 2x is the length after stretching the cord.

So, what is the cord tension in terms of x? (that's x, not Δx)

And u=m/x ( i dont understand why would you use the orignal length of x to obtain the linear mass density instead of the new length of 2x).
No, you're right you should use the new length 2x. But if you use the correct tension, it will work out.

Basically plug that in into the equation waves propagation speed and you get the awnser.

My train of thought for part b ...
I will have to look at part b later, but first let's get part a.
 
  • #4
!! I understand part a completly. I just set it up like df=kdx and did the integral from x to 2x...The math makes more sense now....Thanks!! Sadly iam still stuck in part b.
 
  • #5
Redbelly98
Staff Emeritus
Science Advisor
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You're welcome.

Without having worked it out myself, I believe you'll need to express T and u as functions of time.
 

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