1. Simple Harmonic Oscillator - Motion
Equation of motion for SHO.
Motion animation.
Sinusoidal solution and harmonic frequency.
Terminology and summary.
Resonant frequency animation.
Example problems.
Relation between vmax , a ax , and A.
Problem strategy.
Simple pendulum.

2. Equation of Motion Given the following: What is equation of motion?
𝐹=−𝑘𝑥
𝐹=𝑚𝑎
𝑣= 𝑑𝑥 𝑑𝑡
𝑎= 𝑑𝑣 𝑑𝑡
What is equation of motion?
𝑎=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ?
𝑣=𝑎𝑡+ 𝑣 𝑜 ?
𝑥= 1 2 𝑎 𝑡 2 + 𝑣 𝑜 𝑡+ 𝑥 𝑜 ?
Of course not!
Must be something oscillatory!!

3. Compare with Circular Motion
Compare Simple Harmonic and Circular Motion

4. Simple Harmonic vs. Circular Motion
Simple Harmonic vs. Circular Motion Animation
See Appendix for running embedded files.
To view web link put Powerpoint in reading view.
To view embedded file put Powerpoint in normal view.

5. Sinusoidal Solution 1 Try this form: Must solve equation: Reason:
𝐹=𝑚𝑎=−𝑘𝑥
Try this form:
𝑥=𝐴 𝑠𝑖𝑛𝜔𝑡
Reason:
Goes back and forth like animation
sin its own 2nd derivative
extra terms give flexibility
Derivatives
If 𝑥=𝐴 𝑠𝑖𝑛𝜔𝑡
then 𝑣= 𝑑𝑥 𝑑𝑡 =𝐴𝑐𝑜𝑠𝜔𝑡 𝜔
and 𝑎= 𝑑𝑣 𝑑𝑡 =−𝐴𝑠𝑖𝑛𝜔𝑡 𝜔 2

6. Sinusoidal Solution 2 Plug in on both sides Must solve equation:
𝐹=𝑚𝑎=−𝑘𝑥
Plug in on both sides
−𝑚 𝜔 2 𝐴 𝑠𝑖𝑛𝜔𝑡=−𝑘 𝐴 𝑠𝑖𝑛𝜔𝑡
Wonderful things happen
The “sin” terms cancel.
The minus signs cancel
𝑚 𝜔 2 =𝑘 𝜔= 𝑘 𝑚
Result
Both sides track each other (sine or cosine)
Natural resonant frequency 𝜔= 𝑘 𝑚 𝜔=2𝜋𝑓

7. Resonant frequency Resonant frequency Animation
2π𝑓=𝜔= 𝑘 𝑚
Animation

8. Harmonic Oscillator Terminology
Cycle – One complete oscillation
Amplitude – Endpoint limits (x = -A to +A)
Period – Time to complete one cycle
Frequency – Number of cycles per second
Frequency vs. Period
f = 1/T
T = 1/f
𝜔=2𝜋𝑓= 2𝜋 𝑇

9. SHO - summary to date Energy Motion Harmonic frequency
𝐸= 1 2 𝑘 𝑥 𝑚 𝑣 2 = 1 2 𝑘 𝐴 2 = 1 2 𝑘 𝑣 𝑚𝑎𝑥 2
Motion
𝑥=𝐴𝑠𝑖𝑛 𝜔𝑡 𝑜𝑟 𝑥=𝐴𝑐𝑜𝑠 𝜔𝑡
Harmonic frequency
𝜔= 𝑘 𝑚
Frequency and period
𝜔=2𝜋𝑓 𝜔= 2𝜋 𝑇

10. Example – Problem 7 𝜔=2𝜋𝑓 = 2𝜋4 𝑠 = 25.1 𝑠
m, k, ω – if you know 2/3 you can always find 3rd
𝜔=2𝜋𝑓 = 2𝜋4 𝑠 = 𝑠
𝜔= 𝑘 𝑚
𝑘= 𝜔 2 𝑚
𝑘= 𝑠 2 ∙ 𝑘𝑔= 𝑘𝑔 𝑠 2 = 𝑁 𝑚
For m = kg
𝜔= 𝑘 𝑚 = 𝑁 𝑚 𝑘𝑔 = 𝑠 𝑓=2.83 𝐻𝑧

11. Example – Problem 9 (I) 𝜔=2𝜋𝑓 = 2𝜋3 𝑠 = 18.85 𝑠
m, k, ω – if you know 2/3 you can always find 3rd
𝜔=2𝜋𝑓 = 2𝜋3 𝑠 = 𝑠
𝜔= 𝑘 𝑚 𝑘= 𝜔 2 𝑚
𝑘= 𝑠 2 ∙0.6 𝑘𝑔= 𝑁 𝑚
Total Energy – Just find potential at full amplitude
𝐸 𝑡𝑜𝑡 = 𝟏 𝟐 𝒌 𝒙 𝟐 𝑚 𝑣 2 = 1 2 𝑘 𝐴 2 = 𝑁 𝑚 𝑚 2 =1.8 J
Velocity at equilibrium point
1.8 𝐽= 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 2 + 𝟏 𝟐 𝒎 𝒗 𝟐 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 𝑣 𝑚𝑎𝑥 =2.45 𝑚 𝑠

12. Example – Problem 9 (II) Velocity at 0.1 m 𝐸 𝑡𝑜𝑡 =1.8 𝐽
𝑃𝐸= 𝑁 𝑚 𝑚 2 =1.067 𝐽
𝐾𝐸=1.8 𝐽−1.067 𝐽=0.734 𝐽
1 2 𝑚 𝑣 2 = 𝑣=1.56 𝑚/𝑠
Starting condition: at 𝑡=0 𝑥=±𝐴
Must use cosine!
𝑥=𝐴𝑐𝑜𝑠(𝜔𝑡)
𝑥=0.13 𝑐𝑜𝑠(18.85 𝑡)

13. Example – Problem 13 (I) At any point x Amplitude Max velocity
𝐸 𝑡𝑜𝑡 = 𝟏 𝟐 𝒌 𝒙 𝟐 + 𝟏 𝟐 𝒎 𝒗 𝟐
𝐸 𝑡𝑜𝑡 = 𝑁 𝑚 𝑚 𝑘𝑔 𝑚 𝑠 2
𝐸 𝑡𝑜𝑡 =0.056 J J =0.51 J
Amplitude
0. 51 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝐴 𝐴=.06 𝑚
Max velocity
0. 51 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 𝑣 𝑚𝑎𝑥 =.58 𝑚 𝑠

14. Example – Problem 13 (I) Resonant frequency
𝜔= 𝑘 𝑚 = 𝑁 𝑚 3 𝑘𝑔 = 𝑠 𝑓=1.54 𝐻𝑧
Equation of motion?
Since it doesn’t start at either equilibrium or full amplitude, this requires phase angle
We don’t do in this course – to complicated!

15. Example – Problem 19 (I) Oscillation is given in terms of period
𝜔= 2𝜋 𝑇 = 2𝜋 0.65 𝑠 = 𝑠
Starting condition: at 𝑡=0 𝑥=±0.18
Must use cosine!
𝑥=𝐴𝑐𝑜𝑠(𝜔𝑡)
𝑥=0.18 𝑐𝑜𝑠(9.67 𝑡)
Will reach equilibrium after ¼ cycle
𝑡= 𝑠= 𝑠

16. Example – Problem 19 (II) For maximum velocity
𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 𝑚 𝑣 2 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 = 1 2 𝑘 𝐴 2
1 2 𝑚 𝑣 𝑚𝑎𝑥 2 = 1 2 𝑘 𝐴 2
𝑣 𝑚𝑎𝑥 = 𝑘 𝑚 𝐴=𝜔𝐴= 𝑠 𝑚 =1.74 𝑚 𝑠
For maximum velocity (at full amplitude)
𝐹=𝑚𝑎=𝑘𝑥=𝑘𝐴
𝑎= 𝑘 𝑚 𝐴= 𝜔 2 𝐴= 𝑠 𝑚 =16.8 𝑚 𝑠 2

17. Solving SHO problems
If stretched/compressed and release from rest, then you know amplitude and total energy.
If velocity known at equilibrium midpoint, then you know vmax and total energy.
If you know total energy, you can subtract potential or kinetic to get the other.
If you know k and m, you know ω.
𝜔=2𝜋𝑓 𝜔= 2𝜋 𝑇
General form x = A sinωt or x = A cosωt
If oscillation start at equilibrium sine, full-amplitude cosine.

18. vmax , amax , and A Vmax vs. A amax vs. A
1 2 𝑚 𝑣 𝑚𝑎𝑥 2 = 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝐴 2
𝑣 𝑚𝑎𝑥 = 𝑘 𝑚 𝐴=𝜔𝐴
𝑥=𝐴𝑠𝑖𝑛𝜔𝑡 → 𝑣=𝜔𝐴𝑐𝑜𝑠𝜔𝑡 (calculus)
amax vs. A
𝑚𝑎=−𝑘𝑥
𝑎 𝑚𝑎𝑥 =− 𝑘 𝑚 𝑥 𝑚𝑎𝑥 = −𝜔 2 𝐴
𝑥=𝐴𝑠𝑖𝑛𝜔𝑡 → 𝑎= −𝜔 2 𝐴 𝑠𝑖𝑛𝜔𝑡 (calculus)

19. Simple Pendulum From Physics 103 𝐹=−𝑚𝑔 𝑠𝑖𝑛𝜃 For ϴ small and in radians
𝐹≈−𝑚𝑔𝜃
From geometry
𝐹≈−𝑚𝑔 𝑥 𝑙 (𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝐹=−𝑘𝑥)
Resonant frequency is
𝜔= 𝑔 𝑙 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝜔= 𝑘 𝑚
Acceleration is
𝑎=− 𝑔 𝑙 𝑥 (𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑎=− 𝑘 𝑚 𝑥)
Problem 32 (f=0.572 Hz, E = mgl(1-cosθ)

20. Appendix - Animations
I use animations in this course as I think they are helpful. For each animation you can either click on the web link or, if it’s unavailable, click on the embedded file directly.
To use embedded files you have to have the Flash player standalone version loaded. To do this:
Download file flashplayer_11_sa.exe and put it in known location on your computer.
Hit Start menu, type “Control” in the search box, and this will open Control Panel.
Select “Programs”, “Default Programs”, and select “Associate a file type or protocol with a program”.
Wait forever for the darn thing to load.
In the list of file types, scroll down to “.swf”
Highlight “.swf” and hit “Change Program”, then “Browse”
Navigate to where you put flashplayer_11_sa.exe, and hit OK
Finished
For non-ITS computers (store-bought configuration) just click on the file and it will prompt you.
Mobile
The will run on Android, but not the Chrome browser.
I hear there’s a browser for iOS that will also work (Photon Flash Player)