Attenuating floor oscillations with a cushion

[LCSphysicist]

Homework Statement

All below.

Relevant Equations

Trying to do one.

I am trying to find any relation between the three parameters:
Position of the floor wrt an inertial frame f
Position of the cushion wrt floor c
Position of the man wrt cushion m

But this is really confusing, leaving me to a lot of unnecessary variables
Do you know one smart way to start?

 

[haruspex]

The cushion would in general act as a damped spring, where the damping takes the form of a lower spring constant during expansion than during compression. But that's all a bit hard, so just treat it as an ideal spring, constant k.

Let the floor height vary as $A\sin(\omega t)$ and the height of the top of the cushion, relative to the equilibrium position, be y.
Can you write the differential equation? Can you solve it?

 

[LCSphysicist]

The cushion would in general act as a damped spring, where the damping takes the form of a lower spring constant during expansion than during compression. But that's all a bit hard, so just treat it as an ideal spring, constant k.

Let the floor height vary as $A\sin(\omega t)$ and the height of the top of the cushion, relative to the equilibrium position, be y.
Can you write the differential equation? Can you solve it?

I tried to imagine what follows:
The floor is oscillating with the spring, i will make one action by time:
First the spring stretched to y, after the floor will "go up" by $A\sin(\omega t)$ , all this lead me with:

$my'' = -k(y-A\sin(\omega t))$

I think we can ignore the transient solution here, so we end with:

the module of the maximum would need to be equal A/100, solving:

k = m156
Well, when we sit on the spring:

m156x = mg,
x = g/156
x ~ 0.064 m

The answer is: "about to 6 cm"

so i think this is right, the only thing let me skeptical is the fact i need to assume the module of the expression above, this means the spring and the floor are out of phase?

 

[haruspex]

I tried to imagine what follows:
The floor is oscillating with the spring, i will make one action by time:
First the spring stretched to y, after the floor will "go up" by $A\sin(\omega t)$ , all this lead me with:

$my'' = -k(y-A\sin(\omega t))$

I think we can ignore the transient solution here, so we end with:

View attachment 267755

the module of the maximum would need to be equal A/100, solving:

k = m156
Well, when we sit on the spring:

m156x = mg,
x = g/156
x ~ 0.064 m

The answer is: "about to 6 cm"

so i think this is right, the only thing let me skeptical is the fact i need to assume the module of the expression above, this means the spring and the floor are out of phase?

That all looks right.

Wrt phase, seems to me the solution says they are in phase. Of course, in the real world, there would be some damping too, though not proportional to velocity.

When I first looked at this problem I was a bit sceptical, so I did two checks:
- if we double the cushion thickness we would expect double the sinking depth and double the attenuation factor
- dimensional analysis gives the only way to combine the data to get a distance as $g\omega^2$
So this suggests the answer is proportional to $g\omega^2\times$ attenuation, but there could have been some constant factor, like 2 or pi. Turns out it is 1.