Oscillation of a vertical spring

[voko]

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)?

 

[CAF123]

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.

 

[ehild]

'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.

 

[voko]

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.

 

[CAF123]

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?

 

[voko]

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.

 

[ehild]

So L(0) = b if the system is in equilibrium and L'(0) = -cn?

That looks correct.

ehild

 

[CAF123]

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.

 

[voko]

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.

 

[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.

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

 

[ehild]

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