UNIT 6 LESSON 1
SIMPLE HARMONIC MOTION
SHM
Simple Harmonic Motion Objectives:
Period - Frequency
Hooke’s Law
Energy - Dynamics
Simple Harmonic Motion Homework:
Serway Read pages 390 – 394
Page 394 Example 13.1
Page 414 Problems
#’s 1,3,5,6
t
http://ocw.mit.edu/high-school/physics/oscillations-gravitation/simple-harmonic-motion/
MIT shm pend LECTURE on SHM 8 min
http://ocw.mit.edu/high-school/physics/oscillations-gravitation/simple-harmonic-motion/
MIT LECTURE on SHM and POTENTIAL Energy 5min
.
Simple Harmonic Motion
Simple harmonic motion is typified by the motion of a mass on a spring when it is
subject to the linear elastic restoring force given by Hooke's Law.
The motion is sinusoidal in time and demonstrates a single resonant frequency.
OR
x = xmcos(ωt)
y = ymsin(ωt)
Period T = time (sec) for ONE
Frequency
CYCLE T = 2π(m/k)1/2
f = Cycles/time (sec
-1
) Hertz
Spring Constant k = Newton’s /meter
(Hz)
{ F = -kx}
Hooke's Law
One of the properties of elasticity is that it takes about twice as much force to stretch a
spring twice as far. That linear dependence of displacement upon stretching force is called
Hooke's law.
Spring Potential Energy
Since the change in Potential energy of an object between two positions is equal to
the work that must be done to move the object from one point to the other, the
calculation of potential energy is equivalent to calculating the work. Since the force
required to stretch a spring changes with distance,
the calculation. of the work involves an integral.
The work can also be visualized
as the area under the force curve:
Kinetic energy is energy of motion.
The kinetic energy of an object is the energy it
possesses because of its motion.
The kinetic energy* of a point mass m is still
given by
Energy in Mass on Spring
The simple harmonic motion of a mass on a spring is an example of an
energy transformation between potential energy and kinetic energy. In the
example below, it is assumed that 2 joules of work has been done to set the
mass in motion.
The Kinetic Energy is MAXIMUM @ EQUILIBRIUM
The Potential Energy is MAXIMUM @ X or Y maximum
Simple Harmonic
Dynamic Motion
Equations
The velocity and acceleration are given by
.
The total Energy Dynamics for
an undamped oscillator is the sum of its
kinetic energy and potential energy,
which is constant at
Traveling Wave Equations
Sound / Light
Simple Harmonic Motion Homework:
Serway Read pages 390 – 394
Page 394 Example 13.1
Page: 414 Problems
#’s 1,3,5,6
SHM Day 2 –
Quick Labs
http://www.youtube.com/watch?v=eeYRkW8V7Vg
Position – Velocity – Accelerations of a mass on a spring
http://www.youtube.com/watch?feature=endscreen&v=lIPWyY__N2A&NR=1
Potential – Kinetic – Conservation of Energy
Lab AV-E15
Lab AV-E16
Simple Harmonic Motion - Mathematics
Simple Harmonic Motion - Kinematics
Hooks Law Simulation(s) and SHM LAB
http://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html
SHM Day 2 - Review
Questions:
Position
Velocity – Acceleration
Period
Sinusoidal Motion
Harmonic Oscillations
15 min +- Video with Questions / Solutions
http://ia600600.us.archive.org/33/items/AP_Physics_C_Lesson_18/Container.html
Multiple Choice REVIEW Questions SHM
http://www.learnapphysics.com/apphysicsc/oscillation.php
SHM Day 3 – Review Problems
Problem : At what point
during the oscillation of a spring
is the force on the mass greatest?
Recall that F = - kx . Thus the force on the
mass will be greatest when the displacement
of the block is maximum, or when x = ±xm
Problem : What is the period of oscillation of a mass of 40 kg on a spring with
constant k = 10 N/m?
Recall that T = 2π√{m/k} .
Therefore T = 2π√{40m/10} = 4π seconds
Problem :
A mass of 2 kg is attached to a spring with constant 18 N/m. It is then displaced to the
point x = 2 . How much time does it take for the block to travel to the point x = 1 ?
Recall that x = xm cos(ωt)
ω = 2π/ T = √{k/m} = √{18/2} = 3 radian / second
1 = 2 cos(3t)
1 / 2= cos(3t)
ArcCos(1/2) = 3t
1.0472 = 3t  t = 0.3491 sec
SHM Day 2 - Review Problems
Problem :
A 4 kg mass attached to a spring is observed to oscillate with a period of 2 seconds.
What is the period of oscillation if a 6 kg mass is attached to the spring?
Recall that T = 2π√{m/k} .
Therefore k = 4π2 (m/ T 2 )  k = 4π2
Back to with m= 6kg: T = 2π√{m/k}  T = 2.45 seconds
Problem :
A mass of 2 kg oscillating on a spring with constant 4 N/m passes through its equilibrium
point with a velocity of 8 m/s. What is the energy of the system at this point?
From your answer derive the maximum displacement, xm of the mass.
Recall that KE = ½ mv 2 max at Equilibrium  so KE = 64 Joules
Recall that PE = U = ½ kX m 2
 so with conservation of energy
½ mv 2 = ½ kX m 2
64 = ½ (4) X m 2
SparkNotes Editors. “SparkNote on Oscillations and Simple Harmonic Motion.” SparkNotes.com. SparkNotes LLC. n.d.. Web. 18 Feb.
X m 2 = 2*64/4
 X m = √{128/4} = 5.6569 m
2012.