Simple Harmonic Motion Problem 4

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Homework Help Overview

The problem involves a compound system oscillating in simple harmonic motion (SHM) with a mass m falling into a cage of mass M. Participants are discussing the natural frequency of the system and the implications of momentum conservation during the collision between the falling mass and the cage.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the falling mass's speed and the resulting oscillation amplitude. Some question the validity of using conservation of kinetic energy due to the inelastic nature of the collision. Others discuss the implications of the mean position of oscillation and whether it aligns with the collision point.

Discussion Status

There is ongoing exploration of the problem with various interpretations being considered. Some participants have offered insights regarding the assumptions made about the system's equilibrium and the initial conditions. However, no consensus has been reached regarding the correct approach or answer.

Contextual Notes

Participants note that the problem may lack sufficient information to definitively solve it. There are discussions about the initial conditions of the system and how they affect the calculations, particularly regarding the mean position and the dynamics of the falling mass.

  • #31
TSny said:
A neat thing about a mass oscillating on a vertical spring is that if you let x measure displacement from the equilibrium position (rather than the total stretch of the spring)
which I believe I did
then you can forget about gravity in the oscillations.
For the frequency, yes, but whether you can ignore it for amplitude depends on initial conditions. Often, the initial condition is a known displacement from equilibrium, and in that case the amplitude will be that displacement. But here, we know the velocity at eq.pos. As it descends, the spring has to take up both that KE and some extra lost PE. So the stronger the gravity the greater the amplitude.
 
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  • #32
There would be nonzero amplitude even in the case when the mas m had no speed, but just added to M.

ehild
 
  • #33
ehild said:
There would be nonzero amplitude even in the case when the mas m had no speed, but just added to M.

ehild
Yes, but what was that in response to?
 
  • #34
haruspex said:
Yes, but what was that in response to?

it was in response to you, supporting the statement that the amplitude depends on g. :smile: something like (gm/k) sqrt[1+2hk/(g(M+m))]. But your equation in post #27 does not look correct.

ehild
 
  • #35
ehild said:
But your equation in post #27 does not look correct.
ehild
It's horrid alright, but I can't find a flaw. Do you see one?
 
  • #36
haruspex said:
It's horrid alright, but I can't find a flaw. Do you see one?

I do not know how you got it, but it does not look to give the same A as I got, both from energy and from SHM. Can you show your result?

Have you the change of elastic energy properly taken into account?

ehild
 
Last edited:
  • #37
ehild said:
Have you the change of elastic energy properly taken into account?
No, I had that wrong. Thanks!
 
  • #38
Have you got the same result as me in #34?

ehild
 
  • #39
SammyS said:
My "assumption", which you refer to below, is that at the very beginning of the problem, immediately before the mass begins to fall, at that moment the system is at rest and therefore in equilibrium.

You seem to be assuming that also.

I was thinking you are taking t = 0 at the moment when the particle strikes with the cage.
 

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