Transcript Simple Harmonic Oscillator

Simple Harmonic Oscillator - Motion
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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.
Equation of Motion
– Given the following:
• 𝐹 = −𝑘𝑥
• 𝐹 = 𝑚𝑎
• 𝑣=
• 𝑎=
𝑑𝑥
𝑑𝑡
𝑑𝑣
𝑑𝑡
– What is equation of motion?
• 𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ?
• 𝑣 = 𝑎𝑡 + 𝑣𝑜 ?
• 𝑥=
1
𝑎𝑡 2
2
+ 𝑣𝑜 𝑡 + 𝑥𝑜 ?
• Of course not!
• Must be something oscillatory!!
Compare with Circular Motion
Compare Simple Harmonic and Circular Motion
http://www.animations.physics.unsw.edu.au//jw/SHM.htm
Simple Harmonic vs. Circular Motion
• Simple Harmonic vs. Circular Motion Animation
• http://www.animations.physics.unsw.edu.au//jw/SHM.htm
• See Appendix for running embedded files.
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Sinusoidal Solution 1
• Must solve equation:
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
𝑑𝑥
= 𝐴𝑐𝑜𝑠𝜔𝑡 𝜔
𝑑𝑡
𝑑𝑣
= 𝑑𝑡 = −𝐴𝑠𝑖𝑛𝜔𝑡 𝜔2
– then 𝑣 =
– and
𝑎
Sinusoidal Solution 2
• Must solve equation:
𝐹 = 𝑚𝑎 = −𝑘𝑥
• Plug in on both sides
−𝑚𝜔2 𝐴 𝑠𝑖𝑛𝜔𝑡 = −𝑘 𝐴 𝑠𝑖𝑛𝜔𝑡
• Wonderful things happen
– The “sin” terms cancel.
– The minus signs cancel
– 𝑚𝜔2 = 𝑘
𝜔=
𝑘
𝑚
• Result
– Both sides track each other (sine or cosine)
– Natural resonant frequency 𝜔 =
𝑘
𝑚
𝜔 = 2𝜋𝑓
Resonant frequency
• Resonant frequency
2π𝑓 = 𝜔 =
𝑘
𝑚
• Animation
http://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html
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
SHO - summary to date
• Energy
1
2
1
2
1
2
1
2
𝐸 = 𝑘𝑥 2 + 𝑚𝑣 2 = 𝑘𝐴2 = 𝑘𝑣𝑚𝑎𝑥 2
• Motion
𝑥 = 𝐴𝑠𝑖𝑛 𝜔𝑡 𝑜𝑟 𝑥 = 𝐴𝑐𝑜𝑠 𝜔𝑡
• Harmonic frequency
𝜔=
𝑘
𝑚
• Frequency and period
𝜔 = 2𝜋𝑓 𝜔 =
2𝜋
𝑇
Example – Problem 7
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m, k, ω – if you know 2/3 you can always find 3rd
𝜔 = 2𝜋𝑓 = 2𝜋4 𝑠 = 25.1 𝑠
𝜔=
𝑘
𝑚
𝑘 = 𝜔2 𝑚
𝑘 = 25.1 𝑠
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2
∙ 0.00025 𝑘𝑔 = 0.158
𝑘𝑔
𝑠2
= 0.158
For m = 0.0005 kg
𝜔=
𝑘
𝑚
=
0.158
𝑁
𝑚
.0005 𝑘𝑔
= 17.78
𝑠
𝑓 = 2.83 𝐻𝑧
𝑁
𝑚
Example – Problem 9 (I)
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m, k, ω – if you know 2/3 you can always find 3rd
𝜔 = 2𝜋𝑓 = 2𝜋3 𝑠 = 18.85 𝑠
𝜔=
𝑘
𝑚
𝑘 = 𝜔2 𝑚
2
𝑘 = 18.85 𝑠
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∙ 0.6 𝑘𝑔 = 213.2
Total Energy – Just find potential at full amplitude
𝟏
𝟐
1
2
1
2
𝐸𝑡𝑜𝑡 = 𝒌𝒙𝟐 + 𝑚𝑣 2 = 𝑘𝐴2 =
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𝑁
𝑚
1
2
213.2 𝑁 𝑚 0.13 𝑚
2
= 1.8 J
Velocity at equilibrium point
1
2
𝟏
𝟐
1
2
1.8 𝐽 = 𝐸𝑡𝑜𝑡 = 𝑘𝑥 2 + 𝒎𝒗𝟐 = 𝑚𝑣𝑚𝑎𝑥 2
𝑣𝑚𝑎𝑥 = 2.45 𝑚 𝑠
Example – Problem 9 (II)
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Velocity at 0.1 m
𝐸𝑡𝑜𝑡 = 1.8 𝐽
1
2
𝑃𝐸 = 213.2 𝑁 𝑚 0.1 𝑚
2
= 1.067 𝐽
𝐾𝐸 = 1.8 𝐽 − 1.067 𝐽 = 0.734 𝐽
1
𝑚𝑣 2
2
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= .734
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)
Example – Problem 13 (I)
• At any point x
𝟏
𝟏
𝐸𝑡𝑜𝑡 = 𝟐 𝒌𝒙𝟐 + 𝟐 𝒎𝒗𝟐
1
𝐸𝑡𝑜𝑡 = 2 280 𝑁 𝑚 .02 𝑚
2
1
+ 2 3 𝑘𝑔 .55 𝑚 𝑠
𝐸𝑡𝑜𝑡 = 0.056 J + 0.45375 J = 0.51 J
• Amplitude
1
0. 51 𝐽 = 𝐸𝑡𝑜𝑡 = 2 𝑘𝐴2
𝐴 = .06 𝑚
• Max velocity
1
0. 51 𝐽 = 𝐸𝑡𝑜𝑡 = 2 𝑚𝑣𝑚𝑎𝑥 2
𝑣𝑚𝑎𝑥 = .58 𝑚 𝑠
2
Example – Problem 13 (I)
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Resonant frequency
𝜔=
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𝑘
𝑚
=
280
𝑁
𝑚
3 𝑘𝑔
= 9.66
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!
Example – Problem 19 (I)
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Oscillation is given in terms of period
𝜔=
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2𝜋
𝑇
=
2𝜋
0.65 𝑠
= 9.67 𝑠
Starting condition: at 𝑡 = 0 𝑥 = ±0.18
Must use cosine!
𝑥 = 𝐴𝑐𝑜𝑠(𝜔𝑡)
𝑥 = 0.18 𝑐𝑜𝑠(9.67 𝑡)
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Will reach equilibrium after ¼ cycle
1
4
𝑡 = 0.65 𝑠 = 0.1625 𝑠
Example – Problem 19 (II)
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For maximum velocity
1
2
1
2
1
2
1
2
𝐸𝑡𝑜𝑡 = 𝑘𝑥 2 + 𝑚𝑣 2 = 𝑚𝑣𝑚𝑎𝑥 2 = 𝑘𝐴2
•
1
𝑚𝑣𝑚𝑎𝑥 2
2
= 𝑘𝐴2
1
2
𝑣𝑚𝑎𝑥 =
𝑘
𝐴
𝑚
= 𝜔𝐴 = 9.67 𝑠 .18 𝑚 = 1.74 𝑚 𝑠
For maximum velocity (at full amplitude)
𝐹 = 𝑚𝑎 = 𝑘𝑥 = 𝑘𝐴
𝑎=
𝑘
𝐴
𝑚
= 𝜔2 𝐴 = 9.67 𝑠
2
.18 𝑚 = 16.8 𝑚 𝑠 2
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.
vmax , amax , and A
• Vmax vs. A
1
1 2
2
𝑚𝑣𝑚𝑎𝑥 = 𝐸𝑡𝑜𝑡 = 𝑘𝐴
2
2
𝑣𝑚𝑎𝑥 =
𝑘
𝐴 = 𝜔𝐴
𝑚
𝑥 = 𝐴𝑠𝑖𝑛𝜔𝑡 → 𝑣 = 𝜔𝐴𝑐𝑜𝑠𝜔𝑡
• amax vs. A
(calculus)
𝑚𝑎 = −𝑘𝑥
𝑘
𝑎𝑚𝑎𝑥 = − 𝑥𝑚𝑎𝑥 = −𝜔2 𝐴
𝑚
𝑥 = 𝐴𝑠𝑖𝑛𝜔𝑡 → 𝑎 = −𝜔2 𝐴 𝑠𝑖𝑛𝜔𝑡 (calculus)
Simple Pendulum
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From Physics 103
𝐹 = −𝑚𝑔 𝑠𝑖𝑛𝜃
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For ϴ small and in radians
𝐹 ≈ −𝑚𝑔𝜃
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From geometry
𝐹 ≈ −𝑚𝑔
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𝑔
𝑙
𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝜔 =
𝑘
𝑚
Acceleration is
𝑔
𝑙
𝑎=− 𝑥
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(𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝐹 = −𝑘𝑥)
Resonant frequency is
𝜔=
•
𝑥
𝑙
(𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑎 = −
Problem 32 (f=0.572 Hz, E = mgl(1-cosθ)
𝑘
𝑥)
𝑚
Appendix - Animations
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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.
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To use embedded files you have to have the Flash player standalone version loaded. To do this:
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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
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For non-ITS computers (store-bought configuration) just click on the file and it will prompt you.
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Mobile
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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)