Another Approach
You can use a force diagram
to map out the same
situation.As you can see, the net
force in this case is zero,
as it neither moving up
nor down. The force of the
spring, then, must cancel
out with the weight;
kx = mg.Next, imagine that the block is pulled down a distance A and released. The spring
force increases (because the spring was stretched farther); it’s stronger than the
block’s weight, and, as a result, the block accelerates upward. As the block’s
momentum carries it up, through the equilibrium position, the spring becomes less
stretched than it was at equilibrium, so FS is less than the block’s weight. As a
result, the block decelerates, stops, and accelerates downward again, and the up-
and-down motion repeats.
When the block is at a distance y below its equilibrium position, the spring is
stretched a total distance of d + y, so the upward spring force is equal to k(d + y),
while the downward force stays the same, mg.
New Equilibrium
Normally our equilibrium
for the spring (without a
block) would be at the
position y. When we
attach the block, our new
equilibrium point sits at
(d + y). You can just
replace the old equilibrium
point with the new
equilibrium point to make
calculations easier. Just
remember you still have to