1) A mass attached to a vertical spring can be modeled as a simple harmonic oscillator, with the net force equal to Fnet = -ky. 2) If a 2kg mass is pulled down 20m and released from a spring with k=1, its velocity will be 8m/s at a displacement of 16.49m above the resting point as it moves up. 3) At this point, the kinetic energy is 64J, potential energy is 135.96J, and total energy is 199.96J, consistent with the total energy of 1/2kx^2 = 200J for this simple harmonic motion system.

1. Conservation of Energy in
Simple Harmonic Motion
By: Devin Gamble

2. Consider a mass m suspended by
a spring with spring constant k:
 The difference between the
unstretched spring and spring
supporting the mass is ∆L
 We already know that the force of
the spring Fs acts upwards while
gravity acts downwards.
 Describe an equation for the net
force acting on the mass in
equilibrium

3. Question:
 As the mass is pulled down 20 m and
released, at what positive
displacement (above the resting point
y=0) is velocity equal to 8 m/s?
 For your calculations use m = 2kg for
the mass, and 1 for the spring constant
kFs

4. Follow Up Question:
 With the velocity and displacement
from the previous problem, calculate
the Kinetic and Potential energies at
these instances and thus the Total
energy of the system.
 Consider how the Potential and Kinetic
energies in this system compare to
those in other models of simple
harmonic motion, such as a horizontal
spring or a point travelling around a
circle.

5. Solutions
 The net force acting on a mass held by a vertical spring can be
expressed as the elastic force and the force of weight:
Fnet = k(∆L – y) – mg = k ∆L – ky – mg
Fnet = -ky
- Where ∆L – y denotes the change in the spring’s length after the mass
is added. The relation k∆L – mg = 0 can be used to simplify the
expression to Fnet = -ky.

6. Solutions (con’t)
 Question 1: Since the mass is pulled down a certain distance, 20 m will
be the amplitude of the oscillating mass. A positive displacement y will
be when the spring is contracting and above the equilibrium position.
 With the given information, we can use the equation for displacement
under simple harmonic motion.
*Note: since this is a vertical
model we use y instead of x
x = +/- √(A2 – (
𝑚
𝑘
v2))
Since the question specifies a positive displacement we find that
y = 16.49 m. Since the problem also specifies a positive velocity, the time
at this displacement will be when the mass is ascending and the spring is
contracting.

7. Solutions (con’t)
 Follow up Question:
 Using the principle of energy conservation, E = K + U,
E = ½ mv2 + ½ ky2 respectively
K = ½ mv2 = ½ (2)(82) = 64 J
U = ½ ky2 = ½ (1)(16.492) = 135.96 J
Total energy of the system E = about 200 J (199.96 J)
Could also use E = ½ kA2 = ½ (1)(202) = 200 J
 Potential and kinetic energies for systems in simple harmonic motion
will remain relatively similar. A mass on a horizontal spring will still
have a potential energy, though it will depend on a x value instead of
y. A point moving around a circle will have contrasting kinetic and
potential energies with respect to either the x or y axis.