Simple Harmonic Motion Problem 4

[haruspex]

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.

 

[ehild]

There would be nonzero amplitude even in the case when the mas m had no speed, but just added to M.

ehild

 

[haruspex]

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?

 

[ehild]

Yes, but what was that in response to?

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

ehild

 

[haruspex]

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?

 

[ehild]

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

 

[haruspex]

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

No, I had that wrong. Thanks!

 

[ehild]

Have you got the same result as me in #34?

ehild

 

[thunderhadron]

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.