How Long Does a Box Stay In Contact With a Spring as it Bounces Off of it?

[Heyxyz]

Homework Statement

A box weighing 2 kg falls onto a spring with a constant of 300 N/m. How long does the box touch the spring?

Relevant Equations

T = 2π * √(m/k)

T = 2π * √(2/300), T = .513 seconds.

If I divide it by 4/3, I get a final answer of .385 seconds of touch.

I know the box isn't attached for the entire oscillation, so T has to be divided. To me, it makes sense to divide it by 4/3 (when the box falls, the spring is compressed, hits equilibrium, and the launches the box... 3 out of 4 steps), but I've heard that dividing it by 2 is correct. Of course, I could be totally off, and this could be some sort of kinematic problem.

Thank You!

 

[FranzDiCoccio]

My suggestion is:

• think what an entire oscillation would be (remember that in that case the box should remain attached to the spring)
• play a "movie" in your mind, of what is happening to the box. Then play it backwards

This should suggest the correct factor of the period you should take.
I hope it helps

 

[haruspex]

Start off supposing it to be attached and think about the equilibrium position. In a cycle of period T, how long would it spend below the equilibrium position? How does that compare with the loss of contact position in the unattached case?
(I have a feeling the problem does not provide enough information.)

 

[Heyxyz]

Under normal circumstances, would the box fly off one the spring hits equilibrium again, or does it fly off when the spring is extended? That probably play a big role in this.

 

[jbriggs444]

Under normal circumstances, would the box fly off one the spring hits equilibrium again, or does it fly off when the spring is extended? That probably play a big role in this.

When in doubt, state an assumption and work the problem under that assumption.

In this case, I would suggest considering "making contact" and "flying off" to be the points where the spring is compressed just enough so that the acceleration is zero -- the spring is compressed just enough to cancel gravity.

After having solved the problem under that interpretation, you could proceed to worry about the time interval between touching a relaxed spring and compressing the spring enough to reach zero acceleration. Try solving that while letting the initial velocity of the box be an input variable.

 

[FranzDiCoccio]

uhm wait, on second thought: what do you mean exactly by "falls onto a spring"?

I believe what you are thinking "is laid onto a spring" (with zero velocity).

I'm not a native English speaker, but "Falls on a spring" makes me think that the box was released from some height h above the spring.
I did some quick calculations for this latter case, and I think that the answer could be non trivial, and would depend on h.

If, instead, the block is released from h=0 (laid onto the spring with velocity zero), the answer should be easy (as a function of the period).

In this case, I would suggest considering "making contact" and "flying off" to be the points where the spring is compressed just enough so that the acceleration is zero -- the spring is compressed just enough to cancel gravity.

This is a bit confusing. There is just one such (equilibrium) point, and I would not focus on that.
The box makes (and loses) contact with the spring when the latter is not compressed nor extended, i.e. when the elastic force is null, right?
Instead, if the total force is null, the spring is compressed. But the box has been touching the spring for a while, at that point.

 

[jbriggs444]

uhm wait, on second thought: what do you mean exactly by "falls onto a spring"?

I am not the person who set the question. However, the person answering the question must interpret this remark. The interpretation that I would suggest is an impact with an unknown but significant downward velocity.

I believe what you are thinking "is laid onto a spring" (with zero velocity).

No.

I'm not a native English speaker, but "Falls on a spring" makes me think that the box was released from some height h above the spring.
I did some quick calculations for this latter case, and I think that the answer could be non trivial, and would depend on h.

Yes.

If, instead, the block is released from h=0 (laid onto the spring with velocity zero), the answer should be easy (as a function of the period).

Yes.

This is a bit confusing. There is just one such (equilibrium) point, and I would not focus on that.

Two events. One on the way down and one on the way up. Both, of course, at the same elevation.

The box makes (and loses) contact with the spring when the latter is not compressed nor extended, i.e. when the elastic force is null, right?

That interpretation makes the problem difficult to solve since it introduces a dependency on drop height or impact velocity. My suggestion is intended to eliminate that dependency.

Instead, if the total force is null, the spring is compressed. But the box has been touching the spring for a while, at that point.

Yes.

 

[Heyxyz]

Technically, this is problem has multiple parts, and the block lands (as in dropped) on the spring at 5 m/s after falling some unmentioned number of meters. I did not mention that because I thought it it meant nothing (velocity wasn't in the period formula, so I left it out).

 

[jbriggs444]

Technically, this is problem has multiple parts, and the block lands (as in dropped) on the spring at 5 m/s after falling some unmentioned number of meters. I did not mention that because I thought it it meant nothing (velocity wasn't in the period formula, so I left it out).

So a key insight would be that the motion while in contact with the spring is still a case of simple harmonic motion. You just need to figure out what fraction of a complete cycle is spent in contact with the spring.

[If one had fact-acting glue on the bottom of the box, a full cycle would be obtained and maintained]

 

[Heyxyz]

Would that fraction be 1/2? Here's my perspective, it lands, compress, and let's go once it returns to equilibrium. If it were represented graphically, the entire process would be on one side of the x-axis; it's one half of a total period.

 

[Charles Link]

An important factor here, in how the spring and system responds, is that the spring is assumed to be massless. If the spring has a mass, the problem becomes much more complicated. With this assumption, the problem is fairly straightforward.

 

[haruspex]

Would that fraction be 1/2? Here's my perspective, it lands, compress, and let's go once it returns to equilibrium.

No, it would not be 1/2, and the point where it loses contact is not the equilibrium position. In the equilibrium position the compression in the spring equals the weight of the box, so still very much in contact.

the block lands (as in dropped) on the spring at 5 m/s

Please always supply the information you are given, even if you do not see the relevance. This is why I wrote that there was not enough information.

As @jbriggs444 and I have both advised now, start off by treating it as attached. What part of the cycle is exactly the same for the unattached case?

 

[Heyxyz]

Okay.

If the box were attached to the spring, it would go on forever. That makes sense. It also makes sense that there is a point in the cycle where the unattached box and the attached box are the same. It was pretty much established that equilibrium in not that point.

This leaves me with the compressed state (1/4) of a cycle and the stretched state (3/4) of a cycle. At the moment, it doesn't make sense that the compressed state would be the same; it seems like it requires weight. However, the stretched state doesn't seem to require weight. It makes sense that there might be no difference whether or not the block is present when the spring is stretched.

So, am I on the right track when I say it let's go on the stretched spring 3/4 of the way, or shall I go back to the drawing board?

By the way, thanks. It's difficult to wrap my mind around this.

 

[FranzDiCoccio]

That interpretation makes the problem difficult to solve since it introduces a dependency on drop height or impact velocity. My suggestion is intended to eliminate that dependency.

I'm not sure what you are saying. I do not think this is a matter of interpretation.
If there is a dependency on h, I do not think one is allowed to cancel it just by interpreting the question in a different way.
The box touches the spring when it touches it, and that is before reaching the equilibrium point.

Maybe what you are saying is that using the points you suggest makes the calculation of the time easier. I do agree with that.
However, it is not evident straight away that the time you are calculating this way is the same time asked by the problem.

If it is, a reasonable argument should be made. If the argument is correct, the time does not really depend on h.
But I'm not so sure about that.

 

[FranzDiCoccio]

Technically, this is problem has multiple parts, and the block lands (as in dropped) on the spring at 5 m/s after falling some unmentioned number of meters. I did not mention that because I thought it it meant nothing (velocity wasn't in the period formula, so I left it out).

Another suggestion: the "homework equations" you included in the OP are not the whole story.
If you are familiar with it, try to work with energy (kinetic, gravitational, elastic).
The velocity of the box as it lands on the spring would be in one of that.
You should be able to figure out how far the block would compress the spring.
Then, as it has already been suggested, try to figure out what the box would do if it remained attached to the spring.
This way you should identify what is a complete oscillation, and (with symmetry arguments) for what part of that time the box stays attached to the spring.

If you are not familiar with energy, you should be able to work out the maximum compression from the dynamics and kinematics of the spring-mass system.

 

[jbriggs444]

This leaves me with the compressed state (1/4) of a cycle and the stretched state (3/4) of a cycle.

You seem to be just guessing here. I see no justification that it should be 1/4 versus 3/4 instead of something else.

Personally, I like the idea of visualizing the motion on a circle. Simple harmonic motion follows exactly the same pattern as the vertical component of uniform circular motion. If you could just find the right circle...

You can calculate the center point of the circle -- the equilibrium height where the box is not accelerated.
You can calculate the bottom point of the circle -- the farthest down the box can reach.

 

[haruspex]

Okay.

If the box were attached to the spring, it would go on forever. That makes sense. It also makes sense that there is a point in the cycle where the unattached box and the attached box are the same. It was pretty much established that equilibrium in not that point.

This leaves me with the compressed state (1/4) of a cycle and the stretched state (3/4) of a cycle. At the moment, it doesn't make sense that the compressed state would be the same; it seems like it requires weight. However, the stretched state doesn't seem to require weight. It makes sense that there might be no difference whether or not the block is present when the spring is stretched.

So, am I on the right track when I say it let's go on the stretched spring 3/4 of the way, or shall I go back to the drawing board?

By the way, thanks. It's difficult to wrap my mind around this.

Write the general equation for the SHM phase of the motion. Take the equilibrium position as height 0.
You have a number of facts you can use to determine the parameters.
At time 0, when contact is first made, you know the velocity. Can you also work out where that is in relation to the equilibrium position?

 

[Charles Link]

This one is a little more complicated than my first impression of it. It is straightforward though if you solve the complete second order differential equation for the motion. Otherwise, an analysis of it gets a little complicated. I don't know whether S.H.O. equations by themselves will give an obvious answer. $\\$ Edit: Using the solution to the differential equation, in the form $x(t)=x_o+C \cos(\omega t-\phi)$, the point of release happens at $\omega t=2 \phi$, which is dependent on the initial velocity. (PF rules keep me from writing out the complete solution=at least until after some further discussion). If I computed this correctly, there is no cut and dry answer that the time is $T/2$ or $T$ or $\frac{3}{4} T$, etc.

 

[Dr.D]

The way to work this problem is to treat the box as attached to the spring after impact, and determine the force at the connection. When that force goes to zero, the two separate.

 

[Charles Link]

Please see my "Edit" to post 18. If I solved it correctly, and I believe I did, this one is interesting.

 

[Charles Link]

$y_{sep}$, i.e. $x_{sep}$ is the relaxed post of the spring, which is $x=0$. Please look over my post 18. Let me write out the differential equation here: $m \ddot{x}=mg-kx$. You get a particular solution which is $x_p=mg/k$ , plus a homogeneous solution $x_h=A \cos(\omega t)+B \sin(\omega t)$, with the total solution subject to $x=0$ at $t=0$, and $\dot{x}=v_o$ at $t=0$.

 

[kuruman]

$y_{sep}$ is the relaxed post of the spring. Please look over my post 18.

I know, I agree with you 100% but I clicked "Post" instead of "Preview" prematurely so I deleted my post to which your #21 is the super fast reply. Anyway, yes the answer is speed-dependent. It would probably be more appropriate to find the critical initial speed of the block at contact so that it barely separates from the spring at the top of the oscillation.

 

[FranzDiCoccio]

I believe there is no need of a differential equation.
The point of maximum compression can be determined via the conservation of energy.
Then you can calculate how far that is below the center of the oscillation, i.e. the equilibrium point.
Then you know that, if the box was attached to the spring, it would reach a point that is at the same distance above the equilibrium point (and above the "impact point" where the spring is at rest).
For symmetry reasons you expect that the box looses contact with the spring where it first touched it. Indeed the spring is not compressed nor extended, and it's not really connected to the box. So it stops there, while the box keep going upwards.

Working with the fact that the simple harmonic motion is a projection of a uniform circular motion should allow to evaluate what is the fraction of the period we are looking for.
That is the fraction of circumference that starts at the impact point and goes back to it. Simple trigonometry should allow to find this fraction, which is the same fraction for the relevant angle.
Sorry, my English is fading away...

 

[Charles Link]

@FranzDiCoccio Very good. Your analysis is equivalent to doing some algebra and analyzing the $\cos(\omega t)$ and $\sin(\omega t)$ terms of the solution to the differential equation. For a class that has not yet had the differential equations, your method would be the way to proceed.