Simple Harmonic Motion Problem 4

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SUMMARY

The discussion centers on a Simple Harmonic Motion (SHM) problem involving a mass m falling into a cage with mass M, where the oscillation frequency is determined by the formula √[k/(m + M)]. Participants analyze the conservation of momentum during the collision and the implications for the amplitude of oscillation. The correct amplitude is derived as A = √[k/(M + m)] * m√(2gh)/(M + m), but participants identify errors in their calculations and assumptions about the system's equilibrium. Ultimately, the consensus is that the problem's wording may lead to confusion regarding the initial conditions and the mean position of oscillation.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM) principles
  • Knowledge of conservation of momentum and energy
  • Familiarity with oscillation frequency calculations
  • Basic mechanics involving gravitational potential energy
NEXT STEPS
  • Review the derivation of oscillation frequency in SHM using k and m values
  • Study the conservation of momentum in inelastic collisions
  • Learn about energy conservation principles in mechanical systems
  • Explore the implications of equilibrium positions in oscillatory motion
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and oscillatory motion, as well as anyone solving complex SHM problems involving multiple masses and energy conservation.

  • #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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