The document describes a vertical spring-mass system that undergoes simple harmonic motion when displaced from its equilibrium position and released. When the spring is stretched by 20 cm, the net force needed in the downward direction is 1.58 N. Simple harmonic motion occurs because the net force on the mass is directly proportional to the displacement, following Hooke's law, where the force is the negative of the spring constant multiplied by the displacement.

1. MASS SPRING SYSTEMS & SIMPLE HARMONIC
MOTION
The diagram above shows a vertical springmass system. SpringA shows an
un-stretched spring. SpringB shows a spring, which has a mass, attached to
one end; this causes the springto stretch by ∆L. SpringC is a spring that has
been stretched to a displacement, y. If a mass is attached followed by
stretching the spring, then releasing it and allowingit to return to
equilibrium,
a) Determineif this is a simpleharmonic motion.
b) Calculate the valueof the constant, k.
Given: y = 10.0cm
M = 5.00 kg
T = 5.00 seconds
c) Calculate what the net forcein the y direction is, in order to stretch the
spring20.0 cm.
d) What direction is this forceacting in?
A
C
B
∆L
𝑦
Ekam Romana
22292149

2. SOLUTION AND EXPLANATION
a) You know that at equilibrium (B), the spring exerts an upward force on the mass,
known as Hooke’s Law. This is defined as Fspring= - kx. From the diagram given, you
realize that ∆L = x, therefore Fspring= - k∆L. Furthermore, you know that this upward
force is balance by the weight of the mass, which can be defined as w = mg. Equating
these equations you get:
k∆L = mg  k∆L- mg = 0
Now, you can need to account for the displacement caused when the spring was
stretched. The displacement, y, is subtracted from ∆L in order to find the change in
length compared to the unstretched spring. You can now re-write the equation as:
Fnet,y = k∆L- mg
Fnet,y = k(∆L-y)- mg
Replace ∆L with mg/k (calculated from k∆L = mg)
Fnet,y = k(𝑚𝑔/𝑘 -y)- mg
Simplifying the above equation gives you:
Fnet,y = (𝑚𝑔 − 𝑘y) - mg
Fnet,y = − 𝑘y
From this you recognize that in order for a motion to be a simple harmonic motion the
force must be proportional to the displacement by a constant
F= - constant ∙ X
The net force in the y direction is in the same form; therefore this motion is a simple
harmonic motion.
b) mw2 = k  (5) (2 𝜋 / 5)2
= (5) (4𝜋2
/ 25)
= (4𝜋2
/ 5)
c) Fnet,y = − 𝑘y
= − (
4𝜋2
5
)(0.200)
= − 1.58 N  1.58 N of force needed.
d) Downwards, because you need to pull the spring down in order to stretch it.