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Unit 6 Lesson 1 Simple Harmonic Motion SHM

Simple Harmonic Motion Homework: Serway Read pages 390 – 394 Page 394 Example 13.1 Page: 414 Problems #’s 1,3,5,6. Simple Harmonic Motion Objectives: Period - Frequency Hooke’s Law Energy - Dynamics. Unit 6 Lesson 1 Simple Harmonic Motion SHM.

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Unit 6 Lesson 1 Simple Harmonic Motion SHM

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  1. Simple Harmonic Motion Homework: Serway Read pages 390 – 394 Page 394 Example 13.1 Page: 414 Problems #’s 1,3,5,6 Simple Harmonic Motion Objectives: Period - Frequency Hooke’s Law Energy - Dynamics Unit 6 Lesson 1 Simple Harmonic Motion SHM http://ocw.mit.edu/high-school/physics/oscillations-gravitation/simple-harmonic-motion/ MIT LECTURE on SHM 8 min http://ocw.mit.edu/high-school/physics/oscillations-gravitation/simple-harmonic-motion/ MIT LECTURE on SHM and POTENTIAL Energy 5min

  2. . 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) T = 2π(m/k)1/2 Period T = time(sec) for ONE CYCLE 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 motionof 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 MAXIMUM@ EQUILIBRIUM • The Potential Energy is MAXIMUM@ X or Y maximum

  6. Simple Harmonic Dynamic Motion Equations The velocityand accelerationare given by . The total Energy Dynamicsfor 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 http://www.youtube.com/watch?feature=endscreen&v=lIPWyY__N2A&NR=1 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 http://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html

  9. SHM Day 2 - ReviewQuestions: Position Velocity – Acceleration Period Sinusoidal Motion Harmonic Oscillations 15 min +- Video with Questions / Solutions http://ia600600.us.archive.org/33/items/AP_Physics_C_Lesson_18/Container.html Multiple Choice REVIEW Questions SHM http://www.learnapphysics.com/apphysicsc/oscillation.php

  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 = xmcos(ωt) ω = 2π/ T = √{k/m} = √{18/2} = 3 radian / second 1= 2cos(3t) 1/ 2= cos(3t) ArcCos(1/2) = 3t 1.0472 = 3t  t = 0.3491 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/ T2)  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 kXm2  so with conservation of energy ½ mv 2 = ½ kXm2 64 = ½ (4) X m2 X m2= 2*64/4  X m = √{128/4} = 5.6569 m SparkNotes Editors. “SparkNote on Oscillations and Simple Harmonic Motion.” SparkNotes.com. SparkNotes LLC. n.d.. Web. 18 Feb. 2012.

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