Mass on a tensioned cable with axial/transverse oscillations

[jobu]

Homework Statement

This problem isn't actually assigned homework for me, but I wanted to see if I get the concept. Any feedback and corrections to my answer would be appreciated!

A cable is stretched between two anchors, and a small mass is attached to the cable, through the center of the mass, at the cable midpoint. The cable is made of a linear, elastic material. Gravity is not a factor in this model.

1) the mass is pulled down and released, so that a transverse oscillation in created in the cable. Will increasing the tension in the cable in crease the oscillation frequency of the mass, Why or why not?


2) the mass is pulled towards one anchor and released, creating an oscillation along the axis of the cable. What effect will increasing tension have on the oscillation frequency?

Homework Equations

cable acts like a linear spring - Tension T = k*x
k is the spring constant of the cable
x is the difference in the length of the stretched cable versus the unstretched cable (no tension).

The Attempt at a Solution

for the first part, I think that the increase in tension WILL increase the frequency of the mass oscillation. Whenever the mass moves out of line with the anchored cable endpoints, a vertical component of the tension accelerates the mass back towards that axis. Increasing tension in the cable will increase that vertical force component, and the acceleration on the mass. The mass will change velocity more quickly, and so complete the oscillation cycle more quickly ... increasing the frequency.

For the axial oscillation, the increase in tension will have no effect on the mass oscillation. Since tension acts towards both anchors with equal magnitude, any change in cable tension will also increase the forces acting on the mass towards each anchor equally. The changes in force cancel each other out, and the mass oscillates at the original frequency.

Is that reasoning correct?

Thanks!

 

[jobu]

also, one of the assumptions in the problem is that there is no damping.

For the second part of the question, I think that the force moving the mass would be equal to 2*the displacement of the mass from the cable midpoint if the cable acted as a regular spring. But since you can't really compress cables, would the cable part that the mass is moving towards provide any resistance? If not, then I think the force moving the mass would be equal to 1*its displacement fromt he cable midpoint. Does that make sense?

Thanks!

 

[AlephZero]

Is that reasoning correct?

Yes, you got the right conclusions for the right reasons.

also, one of the assumptions in the problem is that there is no damping.

A small amount of damping has a very small effect on the frequency. For "real life" situations like this, the frequency change caused by damping would be negligible compared with other sources of uncertainty in the model.

For the second part of the question, I think that the force moving the mass would be equal to 2*the displacement of the mass from the cable midpoint if the cable acted as a regular spring. But since you can't really compress cables, would the cable part that the mass is moving towards provide any resistance? If not, then I think the force moving the mass would be equal to 1*its displacement fromt he cable midpoint.

Remember the cable is "stretched" so it has an initial tension. Therefore you can "compress" it relative to the starting conditions, by making the tension a smaller positive value.

If the tension in one half of the cable did become zero, the frequency of the oscillation would increase because the stiffness of the cable would be halved (only one side of the cable would be acting like a spring).

Of course if the amplitude of the oscillation was large, you could have a situation where both cables were in tension for the central parts of the oscillation but only one was in tension for the other parts. The mass would then not be moving in SIMPLE harmonic motion, but the period of oscillation would be longer than if both cables were always in tension.

 

[jobu]

Thanks for the explanation, AlephZero. It's really helpful!

It seems like the period of oscillation would increase if the stiffness were decreased... is that what you meant here?

If the tension in one half of the cable did become zero, the frequency of the oscillation would increase because the stiffness of the cable would be halved (only one side of the cable would be acting like a spring).

I guess one part that still confuses me is the effect of the tension if I am treating the cable as a spring ... or actually two springs with one on either end of the mass. If the mass is at rest on the tensioned cable, and then I apply a force to move it towards one of the anchors, is the force equal to 2*k(stiffness)*x(displacement) ?

I know that if the model were actually using two springs, and there was no initial tension, that the answer would be 2*k*x because one spring would be pulling against the movement of the mass and the other would be pushing against it. So the spring reaction forces would be working together against the movement of the mass.

But if the springs were initially under tension, and then I move the mass, it seems strange that the answer would still be 2*k*x, since one spring would be pulling against the movement of the mass and the other spring would be pulling with it. When I wrote out the force equation, though, it seemed to confirm that answer

_____T<------------|==mass==|------------>T
T +k*x<----------------|== F->==|--------->T-k*x , |--> x

F+(T-k*x)-(T+k*x) = 0, F=2*k*x

am I missing something here?

 

[AlephZero]

Yes, that looks right.

Here's a different way to think about it. Start with the cables not connected to the mass. There will be a equal gaps between the cable and the mass on each side. Call size of the gap G.

Think about the right hand cable first. When you stretch it to attach it to the mass, the tension in the cable will be kG. Call that initial tension T.

Now if the mass moves x to the right, the stretch in the cable becomes G-x and the tension is k(G-x) = T - kx (so long as x < G, otherwise the cable would go slack and tension would be zero).

Similarly for the left hand cable, the tension is -k(G+x) = -T + kx

So the resultant force on the mass is 2kx, as you said, and it does not depend on the size of G or T.

 

[jobu]

thanks again for taking time to help me and for the clear explanations, AlephZero