Oscillation of a vertical spring

In summary: The top end's motion is not just its acceleration. It is also its displacement. And its acceleration is not just the acceleration of its acceleration. It also includes the acceleration of its displacement.z is the top end's position. Then x(t) = z(t) -...Ok, I think I see what you mean. So then I have ##\ddot{L} = \ddot{x} - g = \ddot{z} - g - g##, where I am assuming the z is the displacement of the top end from its equilibrium position and the double g is the two terms that come from the acceleration of the top end's displacement. Then the equation is ##m\ddot{L
  • #36
ehild said:
See picture. The ceiling is the zero of coordinate y and downward is the positive direction. The top of the spring is at x=csin(nt). The length of the spring is L. so y=x+L.

For he mass m, [itex]m\ddot {y}=mg-k(L-a)\rightarrow \ddot {x} +\ddot L =g-\frac{k}{m}(L-a)[/itex]
If downwards is positive, shouldn't that be -(-k(L-a))? So this means that L is the extension of the spring from a. This is clear from your diagram.

Many thanks.
Where was it that I was going wrong? Was I too concentrated with the displacement of the bottom mass?
 
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  • #37
CAF123 said:
Consider the spring end to be at a distance -a. Now consider some displacement of the bottom end to -x. This means from the springs natural length, it has extended a distance
(-x +a). But there is also some displacement at the top end by -z. I believe these are related by the length of the spring: (-x + a) + z = l, so x = z + a -l.

The relation between x and z, if downward is negative, is L = z - x. The extension from the natural length is X = -(L - a). The restoring force (at the bottom) is -kX = k(L - a).
 
  • #38
CAF123 said:
If downwards is positive, shouldn't that be -(-k(L-a))? So this means that L is the extension of the spring from a. This is clear from your diagram.

L is the legth of the spring. a is the unstretched length . The change of length is L-a>0. The spring force points upward, so it is -k(L-a), opposite to mg.

Well, the zero is a bit deeper then the ceiling, so the upper part of the spring can move about it.

CAF123 said:
Many thanks.
Where was it that I was going wrong? Was I too concentrated with the displacement of the bottom mass?

I don not quite follow your notations. If you can attach a sketch... Choose a fixed point and give all positions with respect to that point.

ehild
 
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  • #39
I understand where I went wrong now - The question says we have a displacement at the top end and somehow I simply neglected that and drew a sketch with the mass at the end displaced instead. Thank you voko and ehild.

I now need to find the particular solution to this eqn. From the words of the question, should I let the initial conditions be ##L(0) = a## and ##\dot{L}(0) = nc##?

My method was to split up and solve two eqns:
1): Solve ##\ddot{L} + \omega^2 L = \omega^2b \Rightarrow L =b ##by inspection.
2): Solve ##\ddot{L} + \omega^2L = cn^2 \sin(nt)##
I found a solution of ##L = \frac{cn^2}{\omega^2 - n^2} \sin(nt) ##

Putting these together:
General solution to inhomogeneous eqn: General soln to homogeneous eqn + Particular solution to inhomogeneous.
So I get $$L(t) = A \cos (\omega t) + B \sin(\omega t) + b + \frac{cn^2}{\omega^2 - n^2}$$ A and B can be determined by the Initial conditions, but I want to check they are correct before I start.
Thanks.
 
  • #40
The sin(nt) factor is missing from the final expression.

[tex]L(t) = A \cos (\omega t) + B \sin(\omega t) + b + \frac{cn^2}{\omega^2 - n^2} \sin(nt) [/tex].

Good job!

ehild
 
  • #41
ehild said:
The sin(nt) factor is missing from the final expression.

[tex]L(t) = A \cos (\omega t) + B \sin(\omega t) + b + \frac{cn^2}{\omega^2 - n^2} \sin(nt) [/tex].

Good job!

ehild

Thanks. Are the IC correct?
 
  • #42
There are no initial conditions specified. The formula for L(t) is the general solution if ω≠n. You can fit any initial condition to it.
There is one problem left: what happens if ω=n.

ehild
 
  • #43
ehild said:
There are no initial conditions specified. The formula for L(t) is the general solution if ω≠n. You can fit any initial condition to it.
There is one problem left: what happens if ω=n.

ehild
I have the question
4) 'Write down the initial conditions based on the description above (that is the the problem statement) and find L(t).'

I must be able to make some sort of IC from the fact it is moving downwards at t=0, no?

Also if ##\omega = n##then the last term blows up. So the displacement of the top end is in sinc with that of the natural freq of the spring. Hence if this is the case, is it right to say that we can reach infinite displacements of the spring? (I.e the resonance)
 
  • #44
CAF123 said:
I have the question
4) 'Write down the initial conditions based on the description above (that is the the problem statement) and find L(t).'

I must be able to make some sort of IC from the fact it is moving downwards at t=0, no?

You used this assumption when deciding what sign you should have in front of c sin nt. At this stage you have to assume that x(0) = 0 and x'(0) = 0 and derive the conditions for L from that.

Also if ##\omega = n##then the last term blows up. So the displacement of the top end is in sinc with that of the natural freq of the spring. Hence if this is the case, is it right to say that we can reach infinite displacements of the spring? (I.e the resonance)

What happens is that the solution is no longer valid. You need to find a solution that is valid in this case.
 
  • #45
Assuming at t=0 that x(0) =0 and x'(0)=0 would mean that L(0)=a and L'(0) = 0 right? (where x is the displacement of the top end of the spring).

In the case of omega =n, I have:$$L(t) = A \cos (\omega t) + B \sin (\omega t) + b - \frac{c \omega^2}{2 \omega}t\cos (\omega t)$$

What I was trying to say in my last post about the case omega =n was that physically, we can get in theory, infinite amplitude. (I think)
 
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  • #46
"The top of the spring moving downwards at t=0" means that c is positive if the positive direction is downward. You solved a differential equation for L(t). You can use any initial condition, concerning L.. As an example, it can be that initially the spring is stretched by some length d from its original length and released (so dL/dt =0 and L=a+d.) Or the length of the spring is b at t=0 and dL/dt=0.

If ω=n a particular solution is L=-(0.5cn )t cos(nt) + b. Try.

ehild
 
  • #47
ehild said:
"The top of the spring moving downwards at t=0" means that c is positive if the positive direction is downward. You solved a differential equation for L(t). You can use any initial condition, concerning L.. As an example, it can be that initially the spring is stretched by some length d from its original length and released (so dL/dt =0 and L=a+d.) Or the length of the spring is b at t=0 and dL/dt=0.

So, L(0) = a and L'(0) = 0?

If ω=n a particular solution is L=-(0.5cn )t cos(nt) + b. Try.
See my last post
 
  • #48
CAF123 said:
So, L(0) = a and L'(0) = 0?

That is also a possible choice for initial conditions. Do not forget that you have a differential equation for L, length of the spring. The motion of the top is given.

I haven't seen your last post before sending my one. OK , that is the solution when ω=n.


ehild
 
  • #49
ehild said:
That is also a possible choice for initial conditions. Do not forget that you have a differential equation for L, length of the spring. The motion of the top is given.

Actually I reread the question. It explicitly says '...moving downwards at t=0' so does this not imply that the spring may be stretched? ( and if so, L(0) = a would not hold)
 
  • #50
The top of the spring is moving downwards according to the function csin(nt). At t =0 the displacement of the top is zero. The whole spring is below its top, is'nt it? And a mass is attached to the bottom end. Initially you can do something with that mass, lift up, pull down, or giving it some velocity. So the spring can be stretched initially, but can be unstretched. Or it can be even compressed.

ehild
 
  • #51
L = z - x = - c sin nt - x.
L' = - cn cos nt - x'

if x(0) = x'(0) = 0, what are L(0) and L'(0)?
 
  • #52
voko said:
L = z - x = - c sin nt - x.
L' = - cn cos nt - x'

if x(0) = x'(0) = 0, what are L(0) and L'(0)?
Would that give L(0)=0 and L'(0)=-cn?
L(0) being 0 at t=0 does not make much sense to me since the spring is always going to have its natural length.
 
  • #53
'Write down the initial conditions based on the description above (that is the the problem statement) and find L(t).'
A mass m hangs in equilibrium at the lower end of a vertical spring of natural length a, extending the spring to be a length b.
In case of initial setup that the hanging mass is in equilibrium before the top of the spring starts to oscillate, the initial conditions are L(0)=b and L'(0)=0.

ehild

Edit: I meant y'=0, so L'(0)=-cn.
 
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  • #54
CAF123 said:
Would that give L(0)=0 and L'(0)=-cn?
L(0) being 0 at t=0 does not make much sense to me since the spring is always going to have its natural length.

You are right, x(0) is not zero. At t = 0 the system is in equilibrium.
 
  • #55
voko said:
You are right, x(0) is not zero. At t = 0 the system is in equilibrium.

So L(0) = b if the system is in equilibrium and L'(0) = -cn?
 
  • #56
That is how I interpret the formulation:

"A mass m hangs in equilibrium at the lower end of a vertical spring of natural length a, extending the spring to be a length b."

That means x(0) = -b, x'(0) = 0.

"2) The top end of the spring is made to oscillate vertically with a displacement csin(nt), moving downwards at t=0."

The conditions on the length follow directly from the given law of the top end's motion and the equilibrium condition on the bottom end. "Moving downward" here states explicitly that the top end is moving at t = 0, which in conjunction with x'(0) = 0 can only mean that the length is changing at t = 0.
 
  • #57
CAF123 said:
So L(0) = b if the system is in equilibrium and L'(0) = -cn?

That looks correct.

ehild
 
  • #58
ehild said:
That looks correct.

ehild

Okay, the only thing I don't understand now is that in the question it says the top end is moving downwards at t=0. So wouldn't that move the whole system out of equilibrium?
Thanks.
 
  • #59
CAF123 said:
Okay, the only thing I don't understand now is that in the question it says the top end is moving downwards at t=0. So wouldn't that move the whole system out of equilibrium?
Thanks.

The mass, not the entire system, is said to be in equilibrium. Even though that still sounds a little fishy.

The trouble that you have with grasping the behavior is due to a tiny bit of unrealism in the system: it is not possible to make the top end move in STRICT accordance with c sin nt and at the same time have x'(0) = 0. The sine extends infinitely back in time, so there must have been some oscillations in the past to have the motion in the form c sin nt and to be able to take its derivative. And if there were some oscillations, we can't assume x'(0) = 0. If there were none, the law of motion is not sinusoidal, and z'(0) may not even exist.
 
  • #60
CAF123 said:
Okay, the only thing I don't understand now is that in the question it says the top end is moving downwards at t=0. So wouldn't that move the whole system out of equilibrium?
Thanks.

Imagine you have a motor attached to the top of the spring which makes the top end moving. Before starting the motor the mass is in equilibrium, y=b and y'=0. As you switch the motor on, the displacement of the top end is still zero, but its velocity is -cn. So L(0)=b and L'(0)=-cn.

Initially the hanging mass does not "feel" anything, as the spring force is still k(b-a), and it is in equilibrium with gravity.

As time proceeds, the spring gets shorter, the spring force decreases, so the mass starts to descend.

ehild
 
  • #61
See the figure: It shows the time dependence of both the length L and the position of the mass, (y) for the case when w=2pi, n=4pi, c=0.1 b=1.Note how much less the mass moves compared to the spring.

Also watch the video about an experiment with a slinky. The top of the slinky is released and falls down, while the mass at the bottom stays almost motionless except the last stage when the slinky gets relaxed.

http://www.youtube.com/watch?v=oKb2tCtpvNU&NR=1

ehild
 

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