1. 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
MIT shm pend LECTURE on SHM 8 min
MIT LECTURE on SHM and POTENTIAL Energy 5min

2. Simple Harmonic Motion.
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 CYCLE
T = 2π(m/k)1/2
Frequency f = Cycles/time (sec-1 ) Hertz (Hz)
Spring Constant k = Newton’s /meter { F = -kx}

3. .
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.

4. 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

5. 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 EQUILIBRIUM
The Potential Energy is X or Y maximum

6. 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

7. 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

8. SHM Day 2 – Quick Labs http://www.youtube.com/watch?v=eeYRkW8V7Vg
Position – Velocity – Accelerations of a mass on a spring
Potential – Kinetic – Conservation of Energy
Lab AV-E15 Simple Harmonic Motion - Mathematics
Lab AV-E16 Simple Harmonic Motion - Kinematics
Hooks Law Simulation(s) and SHM LAB

9. SHM Day 2 - Review Questions:
Position
Velocity – Acceleration
Period
Sinusoidal Motion
Harmonic Oscillations
15 min +- Video with Questions / Solutions
Multiple Choice REVIEW Questions SHM

10. 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
= 3t  t = sec

11. 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
X m 2 = 2*64/4  X m = √{128/4} = m
SparkNotes Editors. “SparkNote on Oscillations and Simple Harmonic Motion.” SparkNotes.com. SparkNotes LLC. n.d.. Web. 18 Feb