A solid metal disk suspended from a spring is used in a physics problem involving conservation of energy. A stone with mass C falls from height D and sticks to the disk. The disk then undergoes simple harmonic motion (SHM). The summary provides the speed of the stone immediately upon impact, the period of the SHM of the disk+stone system, and the maximum stretch of the spring during oscillation. Adding a lighter bouncy ball would decrease the period of SHM compared to the stone of the same mass.

2. Long-answer question
A solid metal disk is suspended from a spring that is attached
to a flat roof, which extends downwards with a distance Y.
Note that the mass of the metal disk is given as C. A stone
with the same mass descends from altitude D on the disk and
stays attached to it. All answers must be in terms of given
consonants. Diagram is given below.

3. Diagram of the
situation:

4. a. What is the speed of the rock as soon as it falls into the
disk?
b. Figure out the period T of the SHM that takes place.
c. At the disk’s maximum speed, what will the stretch
distance of the spring be from its initial position?
d. Now the stone is taken out from the top of the disk,
which then restores back to its equilibrium position. A
bouncy ball, with the same mass as the stone, is then
dropped from the same height D, which sticks to the
disk. Will the disk’s SHM period that occurs increase,
decrease, or remain the same?

5. solution
a. As the energy is conserved:
mgh = (1/2)mv^2
CgD = (1/2)Cv(stone)^2
*Cancel out C as the mass remains the same
gD = (1/2)v(stone)^2
*Solve for v(stone)
v(stone) = sqrt(2gD)

6. b. Period of a mass on a spring:
T = 2π(sqrt(m/k))
In this case: m = 2C
To solve for k, use the force equation for
the spring’s initial stretch:
mg = kY, gives k = (mg)/Y
Substitute m and k in the period
equation:
T = 2π(sqrt(2Y/g))
As m cancels out.

7. c. The stretch distance from initial position:
The speed is maximum at equilibrium point:
F = 0, a = 0, Kinetic Energy is maximum,
midpoint of the oscillation.
At new equilibrium point: kx = 2Cg
(x being the distance of the spring stretched
from its initial (unstretched) position.
Substitute k (found in part b.) value:
((Cg)/Y)x = 2Cg
x = 2Y

8. d. SHM period decreases in the case of bouncy ball.
As the bouncy ball is lighter in mass than the
stone of the same mass.
As the mass decreases, the period decreases.
This can be seen by the equation:
T = 2π(sqrt(m/k))