Spring amplitude when glued balls separate at lowest point

[byu]

Homework Statement

A 1.50 kg ball and a 2.00 kg ball are glued together with the lighter
one below the heavier one. The upper ball is attached to a vertical ideal
spring of force constant 165 N/m and the system is vibrating vertically with
amplitude 15.0 cm. The glue connecting the balls is old and weak, and it
suddenly comes loose when the balls are at the lowest position in their
motion. Find the amplitude of the vibrations after the lower ball has come loose.

Homework Equations

ΣF=ma
W=mg
F_s=-kx
m₁=1.5kg
m₂=2kg

The Attempt at a Solution

∑F=ma
-kx₁-(m₁+m₂)g=0
x₁=(m₁+m₂)g/k
x₁= .2079m

.15m+x₁=.3579m

x₂=(m₂)g/k
x₂=.11879m

.3579m - .11879m = .239m

Even after solving it, I do not really understand what is going on in the problem. Why is x=.2079 with both of them attached, when the amplitude is .15m?

 

[paisiello2]

x=.2079 is the displacement of the spring to the equilibrium point before the loosening. We are assuming in both cases that the masses vibrate about the equilibrium point.

 

[byu]

So, it looks like this?

 

[paisiello2]

Yes.

 

[byu]

Thing is; when x₂ = 0.11879m its amplitude is 0.239m. Wouldn't that overshoot into the wall and affect its amplitude?

 

[paisiello2]

Good point but the assumption is that the spring has some minimum length, at least 0.208m. So your diagram should show this datum point and the wall offset that minimum amount.