Physics 104 – Spring 2017 Intro and Harmonic Oscillator Energy Introduction and Syllabus Procedures (same as 103) Topics covered Differences with Phy 103 Harmonic Oscillator Moscow, Russia December 2016

General Physics: Thermodynamics, Electromagnetism, Optics Intro and Syllabus Physics 104 General Physics: Thermodynamics, Electromagnetism, Optics Spring 2016 Syllabus   Instructor: Nat Hager, Research Scientist, Physics and Engineering 173 Masters/Esbenshade (behind mineral gallery) Email: nehager@msi-sensing.com - also forwards to smartphone. (OK to nag if I don’t reply in a day or 2). Web: msi-sensing.com/etown.htm or public directory “hagerne” Phone: Office/Lab 361.1377. Home: 898.3053 before 9:00 PM. Please leave a message. Class Hours: Tue/Thu/Fri 11:00 AM – 12:20 PM Esbenshade 270 All class periods are the same format. Discussion topics will be covered in all sessions. Lab: Thu 2:00 – 3:50 PM or 4:00 – 5:40 PM Office Hours: Tuesday, Friday 1:00 – 2:00 PM Or by appointment. Please feel free to stop by my lab anytime, if my door is closed please leave a note. Prerequisites: Physics 103 or equivalent Textbook: Giancoli, D.C., Physics, Seventh Edition, Prentice Hall, 2014. Supplemental Texts: Many Resources in the Physics Hideaway in Esbenshade including: Boyle, J, Study Guide: Physics (Giancoli), Prentice Hall, 2004.

Procedures Same as Physics 103 WebAssign (setup) Quizzes (5) Exams (3) Lab Final Powerpoints Equation sheets

Topics Waves and Sound (Ch. 11-12) Thermodynamics (Ch. 13-15) Electricity and Magnetism (Ch. 16-22) Optics (Ch. 23) We’re trying to cover a lot, so we’ll have to skip around some. (almost never do starred sections)

General Differences with Physics 103 Use “Energy” in broader context Kinetic energy of ideal gas. (6.02 x 1023 molecules) Potential energy stored in chemical bonds (fuel). Energy stored/carried in EM field. Anything that can do work! Define new Forces Electrostatic 𝐹=𝑞𝐸 Magnetic 𝐹=𝑞𝑣×𝐵 But use same F = ma relations Reinterpret old concepts Gravitational Field g= 9.8 𝑁 𝑘𝑔 (𝑠𝑎𝑚𝑒 𝑎𝑠 𝑚 𝑠 2 ) F=mg Electric Field 𝐸= 𝑥𝑥 𝑁 𝐶 F=𝑞𝐸 Magnetic Field 𝐵= 𝑥𝑥 𝑁 𝐴 ∙𝑚 F=qv×𝐵 Develop analogous methods Flow of fluid (continuity) -> flow of electrical current (Kirchoff)

Simple Harmonic Oscillator A little Physics 103 Jumping-off point for Waves

Simple Harmonic Oscillator Object subject to restoring force around equilibrium: F = - kx Force proportional to and opposite displacement Oscillatory motion around equilibrium Rate determined by mass m and k Frictional damping Examples Block on a spring (car on springs) Meter stick anchored one end (diving board) String in guitar (sound wave) Object bobbing in water Molecule in crystal lattice

Energy in Harmonic Oscillator Potential Energy Work done by relaxing spring: (force x distance) 𝑊= 𝑥 𝑖 𝑥 𝑓 −𝑘𝑥 𝑑𝑥=− 1 2 𝑘 𝑥 𝑓 2 − 1 2 𝑘 𝑥 𝑖 2 Looks like decrease in potential energy Kinetic Energy Energy gained by block being pushed: 𝑊= 1 2 𝑚 𝑣 𝑓 2 − 1 2 𝑚 𝑣 𝑖 2 Appears as increase in kinetic energy Total Energy Loss of potential = gain in kinetic, vice-versa Sum of Kinetic and Potential Constant 𝐸= 1 2 𝑚 𝑣 2 + 1 2 𝑘 𝑥 2 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Limiting points 𝐸= 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 𝐸= 1 2 𝑘 𝐴 2

Harmonic Oscillator Terminology Cycle – One complete oscillation Amplitude – x = -A to x = +A Period – time to make one cycle Frequency – # cycles per second Frequency vs. Period f = 1/T T = 1/f

Example 11-4 - Part 1 Vertical - Find spring constant 𝐹 𝑦 =0 𝐹 𝑦 =0 𝑘 𝑥 𝑜 −𝑚𝑔=0 𝑘= 𝑚𝑔 𝑥 𝑜 = 0.3 𝑘𝑔 9.8 𝑚 𝑠 2 15 𝑚 𝑘=19.6 𝑁 𝑚

Example 11-4 - Part 2 Horizontal - Find total energy 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 2 + 1 2 𝑚 𝑣 2 At maximum amplitude A 𝐸 𝑡𝑜𝑡 = 𝟏 𝟐 𝒌 𝒙 𝟐 + 1 2 𝑚 𝑣 2 = 1 2 𝑘 𝐴 2 𝐸 𝑡𝑜𝑡 = 1 2 19.6 𝑁 𝑚 0.1 𝑚 2 𝐸 𝑡𝑜𝑡 =0.098 J

Example 11-4 - Part 3 At x=0 - maximum velocity vmax 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 2 + 𝟏 𝟐 𝒎 𝒗 𝟐 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 0.098 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 𝑣 𝑚𝑎𝑥 = 2 ∙0.098 𝐽 0.3 𝑘𝑔 =0.81 𝑚 𝑠 At x= 0.05 - velocity 𝐸 𝑡𝑜𝑡 =0.098 𝐽 𝑃𝐸= 1 2 19.6 𝑁 𝑚 0.05 𝑚 2 =.0245 𝐽 𝐾𝐸=0.098 𝐽−0.0245 𝐽=0.0735 𝐽 1 2 𝑚 𝑣 2 =.0735 𝑣=0.7𝑚/𝑠

Example 11-4 - Part 4 Maximum acceleration at maximum stretch 𝑎= 𝑘𝐴 𝑚 𝐹=𝑚𝑎=𝑘𝑥=𝑘𝐴 𝑎= 𝑘𝐴 𝑚 𝑎= 19.6 𝑁 𝑚 0.1 𝑚 0.3 𝑘𝑔 𝑎=6.53 𝑚 𝑠 2

Example – Problem 23 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 2 + 𝟏 𝟐 𝒎 𝒗 𝟐 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 At center point A 𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝑥 2 + 𝟏 𝟐 𝒎 𝒗 𝟐 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 𝐸 𝑡𝑜𝑡 = 1 2 0.755 𝑘𝑔 2.96 𝑚 2 𝐸 𝑡𝑜𝑡 =3.31 J Amplitude 3. 31 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝐴 2 𝐴= 2 ∙3.31 𝐽 124 𝑁 𝑚 =0.231 𝑚

Example – Problem 13 At any point x Amplitude Max velocity 𝐸 𝑡𝑜𝑡 = 𝟏 𝟐 𝒌 𝒙 𝟐 + 𝟏 𝟐 𝒎 𝒗 𝟐 𝐸 𝑡𝑜𝑡 = 1 2 280 𝑁 𝑚 .02 𝑚 2 + 1 2 3 𝑘𝑔 .55 𝑚 𝑠 2 𝐸 𝑡𝑜𝑡 =0.056 J+0.45375 J =0.51 J Amplitude 0. 51 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑘 𝐴 2 𝐴=.06 𝑚 Max velocity 0. 51 𝐽=𝐸 𝑡𝑜𝑡 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 𝑣 𝑚𝑎𝑥 =.58 𝑚 𝑠

Vertical Harmonic Oscillator

Summary - Harmonic Oscillator Energy At any position 𝐸= 𝟏 𝟐 𝒎 𝒗 𝟐 + 𝟏 𝟐 𝒌 𝒙 𝟐 At full amplitude: 𝐸= 1 2 𝑚 𝑣 2 + 𝟏 𝟐 𝒌 𝒙 𝟐 = 1 2 𝑘 𝐴 2 At max-velocity midpoint: 𝐸= 𝟏 𝟐 𝒎 𝒗 𝟐 + 1 2 𝑘 𝑥 2 = 1 2 𝑚 𝑣 𝑚𝑎𝑥 2 Same for all 3: find for one case, know it for all. Find total energy, find one component, subtract for other.

Harmonic Oscillator- Tip Do not waste time memorizing this stuff: 𝑣=± 𝑣 𝑚𝑎𝑥 1− 𝑥 2 𝐴 2 You don’t need it You’ll soon forget it. It trivializes the underlying physics Instead Learn 𝐸=𝐾𝐸+𝑃𝐸 and how KE and PE just get swapped back and forth!