1. Physics 2 – May 3, 2018 P3 Challenge –
What is the Q factor for the following underdamped oscillation? (The scale represents cm.)

2. Objectives/Agenda/Assignment
9.1 Simple Harmonic Motion – Revisit
Agenda:
Derivation of Cosine/sine modeling for oscillations
Period of mass on a spring
Period of pendulum
Energy of oscillation including cosine modeling

3. Rotational speed….Review
One way to determine how fast something is spinning is to count its frequency of rotation or revolutions per second or minute, or rpm.
Given the symbol f, just like frequency for oscillations or waves.
A related idea is the period of the revolution, T which is the reciprocal of f.
Another measure is the angular frequency , ω, (lower case Greek letter omega)
Unit for angular frequency is radians/sec
Period is also related to ω: 𝑻= 𝟐𝝅 𝝎
Angular frequency and frequency are also related: 𝝎=𝟐𝝅𝒇

4. Circular motion and SHM
If you consider a shadow cast by a rotating object as it rotates in front of a wall, you will see the shadow oscillate up and down with Simple Harmonic Motion.
Therefore, there is a one to one correspondence or functional relationship between circular motion and SHM.
The shadow SHM displacement represents the x or y component of the circular motion
X component is r cos θ and Y component is r sin θ
θ = ωt (circular version of distance = rate times time.)
r is the amplitude of the oscillation, xo
Putting this together, x = xo cos ωt or x = xo sin ωt
This is how SHM displacement is modeled with sine curves.

5. Deriving from sine/cosine model
Consider Graph of x = xo cos ωt Displacement function of time.
Amplitude = xo
Slope of displacement graph = velocity
Take the derivative v = – ω xo sin ωt vmax = ω xo
Graph of v = – ω xo sin ωt Velocity function of time.
Slope of velocity graph = acceleration
Take the derivative again a = – ω2 xo cos ωt
Substitute a = – ω2 x amax = ω2 xo
Meets the condition for the acceleration to be proportional to and opposed to displacement! …. A restoring force
General case with phase derived
From Data Packet
𝜔=2𝜋/𝑇
𝑎=−𝜔2𝑥
𝑥=𝑥0sin𝜔𝑡; 𝑥=𝑥0cos𝜔𝑡
𝑣=𝜔𝑥0cos𝜔𝑡; 𝑣=−𝜔𝑥0sin𝜔𝑡

6. Mass on spring oscillation
Consider a = – ω2 x
Also know from Hooke’s Law and the 2nd law, F = – kx = ma
Substitute for a – kx = m(– ω2 x)
Simplify k = m ω2  useful equation (not in data booklet)
Solve for ω, substitute into T = 2π/ ω gives equation in data booklet for period.
𝑇=2𝜋 𝑚 𝑘

7. Pendulums Another restorative force creating an oscillation.
The net force = – mgsinθ
θ = s/L angular displacement in radians
Is net force proportional to and opposite to θ??
F = – mg sin (s/L) = – mg (s/L)
sinx ~ x only at small angles. (compare sine curve to y=x)
Therefore, for pendulums with small angles of displacement we can use the generic cosine/sine modeling for oscillation

8. Pendulums F = ma = – mg (x/L) a = – g (x/L) for small angles
a = – ω2 x From general oscillations
Therefore, – g (x/L) = – ω2 x x cancels for all values of x
g/L = ω2
Solve for ω, substitute into T = 2π/ ω gives equation in data booklet for period.
𝑇=2𝜋 𝐿 𝑔

9. Problems
1. What is the length of a pendulum that has a period of second?
Calculate the percentage increase in the period of a pendulum when the length is increased by 4.00%. What is the new period?
2. A spring is brought to the international space station and used as a balance. If a kg mass oscillates with a sec period, what is the mass of an object that oscillates 13 times in 10 seconds when suspended from the spring?

10. Energy considerations from Ch 4
What we already know about Energy
ET =EK + KP = ½ mv2 + ½ kx2 for a mass on a spring
ET = ½ mvmax2 = ½ kxmax2 for a mass on a spring
𝒗 𝒎𝒂𝒙 = 𝟐 𝑬 𝑻 𝒎
Kinetic E is zero at extremes and maximum at equilibrium
Potential E is maximum at extremes and zero at equilibrium
Graph of a vs x is a line with a negative slope.
Extremes of the graph represent the amplitude

11. Energy with Cosine modeling
Kinetic EK = ½ mv2
𝒗=−𝝎 𝒙 𝒐 𝒔𝒊𝒏(𝝎𝒕)
Substituting: EK = ½ m(𝝎 𝒙 𝒐 𝒔𝒊𝒏(𝝎𝒕))2
EK =1/2 mω2 xo2sin2(ωt)
ET = ½ mvmax2
vmax = ω xo from cosine/sine model
Substituting: ET = 1/2 mω2 xo2

12. Energy with Cosine modeling
Potential EP = ET – EK Because sum of kinetic and potential = total
Substituting: EP = 1/2 mω2 xo2 – 1/2 mω2 xo2sin2(ωt)
Factor out 1/2 mω2 xo2 : EP = 1/2 mω2 xo2 (1 – sin2(ωt))
Substitute from trig identity: cos2(ωt) = (1 – sin2(ωt)
EP = 1/2 mω2 xo2 cos2(ωt)

13. Energy Summary (E as function of t)
EK =1/2 mω2 xo2sin2(ωt)
EP = 1/2 mω2 xo2 cos2(ωt)
ET = 1/2 mω2 xo2
Notice: These apply for every kind of oscillation
Sine2 and cosine2 functions, as expected
Notice Period is two oscillations of energy

14. Velocity formula
EK =1/2 mω2 xo2sin2(ωt) = ½ mv2 Set EK Oscillation = ½ mv2
Cancel ½ and m and switch sides
v2 = ω2 xo2sin2(ωt)
Substitute trig identity sin2(ωt) = (1 – cos2(ωt))
v2 = ω2 xo2 (1 – cos2(ωt))
Distribute: v2 = ω2 xo2 – ω2 xo2 cos2(ωt)
Recognize x = xo cos(ωt) to substitute an x2
v2 = ω2 xo2 – ω2 x2
Solve for v: Note +/- 𝒗 = ±𝝎 𝒙 𝒐 𝟐 − 𝒙 𝟐 At any given displacement, the object may going up or down (or back and forth) at that spot, so +/- v

15. Energies as functions of x
Using the velocity formula we can find expressions for EK and EP as function of position.
From the graphs we know both should be quadratic functions in x.
EK =1/2 mv and v2 = ω2 (xo2 – x2)
EK =1/2 mω2 (xo2 – x2) …..so that’s Kinetic
ET = 1/2 mω2 xo2 ….total is still the same
EP = ET – EK = 1/2 mω2 xo2 – 1/2 mω2 (xo2 – x2)
Distribute and cancel….
EP = 1/2 mω2 x2

16. Energy Summary (function of x)
EK = 1/2 mω2 (xo2 – x2)
EP = 1/2 mω2 x2
ET = 1/2 mω2 xo2
Notice: These apply for every kind of oscillation
Quadratic functions as expected
Be able to read T and A from either graph.

17. Calculus Info for #7 – Graph interpretation
#7 requires you to read displacement information off a velocity graph.
Velocity from a displacement graph is the slope of the x vs t graph.
Displacement from a velocity graph is the area between the curve and the x axis. (an integral)
For #7, each of the ”¼ of an egg” shape sectors represents a displacement of from 0 to one amplitude.
Recall for an oscillation, the object travels over four of these kinds of amplitudes during one complete oscillation. Verify this fact on the graph of #7.

18. Other Homework help Remember,
For oscillations, the restorative force is proportional to and opposite to position.
Therefore graphs of Force vs displacement or acceleration vs displacement will be linear with a negative slope.
They will be line segments with a domain of 2 times the amplitude of the oscillation.

19. Exit slip and homework
Exit Slip – What are the useful equations you will need to memorize that are not part of the data booklet?
What’s due? (homework for a homework check next class)
p358 #1-6, 7-13
What’s next? (What to read to prepare for the next class)
Read Ch 4.4 p and 9.2 p