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Simple Harmonic Motion
Chapter 13**
Simple Harmonic Motion
• An example of periodic motion in which
the restoring force is linear.
• A simple approximation that can be
applied to many real world scenarios and
allows for a straight forward analysis of
these situations.
• It can be applied to bridges, buildings, and
many other real world objects.
Simple Harmonic Motion
• Position oscillates following sinusoidal
pattern
• Amplitude (A) is the maximum distance
that an object moves from its equilibrium
position. A simple harmonic oscillator
moves back and forth between the two
positions of maximum displacement (A
and –A).
Hooke’s Law Review
• Elastic or Spring Force
• FE = -kx
(also called Fs)
• k is the spring constant (N/m) that is unique to
each spring/elastic
• x is the displacement from the equilibrium
position
• **Note: The – part of the equation is because
we are discussing the restoring force of the
spring. The original force that caused the
movement from equilibrium is the positive force
(Newton’s 3rd Law: equal and opposite forces).
Stationary Mass on a Spring
• Restoring force up:
– Fres = Fe = -kx
• In a vertical spring:
– Fg = -Fres as the force pulling the
spring down is gravitational force
and the restoring force would
therefore pull up.
Simple Harmonic Motion
• Fe = 0
Fe = -kx
Fe = kx
Example 1
• A piece of rubber is 45cm long when a
weight of 8.0N hangs from it and is 58cm
long when a weight of 12.5N hangs from it.
• What is the spring constant of this piece of
rubber?
• *** Remember that x is the displacement
from equilibrium position.
Solution
Check Your Learning
• The spring in a typical Hooke’s Law
apparatus has a spring constant of 15.0
N/m and a maximum extension of 10.0
cm. What is the largest mass that can be
placed on the spring without damaging it?
Solution
SHM
• Since restoring force is not constant, we cannot
use kinematics equations
• Period in SHM only depends on mass and
spring constant (not amplitude or x)
• Formula:
• Where T = period (s)
• m = mass (kg)
• k = spring constant (N/m)
Example 2:
• A vertical spring has a spring constant of
43.2 N/m. If a 135g mass is hung on it
and made to vibrate,
• a) what will the frequency of the vibration
be?
• b) what will its amplitude in this situation in
comparison to when it had no fish on it?
Solution
F= kx
Fg = 43.2x
mg = 43.2x
(0.135)(9.81) = 43.2x
x = 0.0306 m or 3.06cm = Amplitude
Check Your Learning
• When an 80.0kg person climbs into an
1100kg car, the car’s springs compress
vertically by 1.2cm.
– What causes the force on the spring?
– What will be the frequency of vibration when
the car hits a bump?
Solution
Practice
• Worksheet: 1*, 2, 3, 4***, 5*, 6
Elastic Potential Energy
• Work is done to stretch or compress a spring.
• Remember this formula from grade 11:
1 2
E e  kx
2
•
•
•
•
Where E is energy (J)
k is spring constant (N/m)
x is the displacement from equilibrium position
***Remember that x can be replaced with
Amplitude (A) as well!
More Grade 11 Review…
Energy can be written as:
Es (spring energy)
Ee (elastic energy)
Ep (potential energy)
Conservation of Energy
• Energy is conserved. Energy is neither created nor
destroyed.
• Total Energy is made up of kinetic energy and the
spring’s potential energy.
Conservation of Energy
• Assuming there is no energy lost to friction
(ie, we are only dealing with mechanical
energy not Q),
Et = Ek + Ee
Things to Remember…
• When the spring is pulled to its max (or compressed to
its max compression) it has only potential energy (Ee).
• Maximum acceleration occurs at the maximum
compression or extension as it will change directions.
• When released, it will have energy from both kinetic and
potential.
• At its equilibrium position, it will have only kinetic energy
(so max velocity).
• F = ma
another equation that might help 
Example 3:
• A spring with a spring constant of 125N/m
is attached to a 245g mass. The mass is
pulled a distance of 2.5cm from
equilibrium and released.
– What is the total energy of this system?
– What is the maximum speed of the mass?
Solution
Check Your Learning
• A block of mass 0.50kg is placed on a
level, frictionless surface, in contact with a
spring bumper (which has a 100.N/m
spring constant). The spring bumper has
been compressed by 0.30m. The spring,
whose other end is fixed, is then released.
• What is the speed of the block at the
instant when the spring is still compressed
by 0.10m?
Solution
Summary
Practice
• Worksheet: Skip 7
• Hint for 10b: Remember that the bumper
only compresses and goes back to
equilibrium – it does not make a full
period.
• Hint for 11: Remember to use momentum
here!