1. Simple Harmonic Motion – Dynamics and Energy Contents: Dynamics Energy Example Whiteboards

2. Simple Harmonic Motion - Dynamics F = ma F = -kx (Not in data packet) ma = -kx a = -kx m Show SHM with graphs – note that acceleration is opposite x x = x o sin(  t) v = x’ =  x o cos(  t) a = x’’ = -x o  2 sin(  t) TOC SOOOO:  =

3. Simple Harmonic Motion - Energy E k (max) = 1 / 2 mv o 2 E p (max) = 1 / 2 kx o 2 Where they happen TOC E k : 0 max 0 E p : max 0 max Derive the energy equations: E k = 1 / 2 m  2 ( x o 2 – x 2 ) E T = 1 / 2 m  2 x o 2

4. Simple Harmonic Motion - Energy E k = 1 / 2 m  2 ( x o 2 – x 2 ) E T = 1 / 2 m  2 x o 2 TOC E T – Total Energy E k – Kinetic Energy  – “Angular” velocity T – Period of motion x – Position (at some time) v – Velocity (at some time) x o – Max Position (Amplitude) v o – Max Velocity

5. Energy Example – An SHO has an amplitude of 0.480 m, a mass of 1.12 kg, and a period of 0.860 seconds. a) what is its angular velocity? b) What is its total energy? What is the maximum kinetic energy and maximum potential energy? c) What is the kinetic and potential energy when the velocity is 3.00 m/s? d) what is the kinetic energy when it is 0.230 m from equilibrium? What is its potential energy here? e) write possible equations for its position and velocity ω = 2π/(0.86 s) = 7.306 ≈ 7.3 rad/sec E k(max) = ½mω 2 x o 2 = ½(1.12 kg)(7.306 rad/sec) 2 (0.48 m) 2 = 6.887 ≈ 6.9 J When it has its maximum kinetic energy, it is passing through the middle (equilibrium) so it has no potential energy, so the total energy is the same as the maximum kinetic energy and the maximum potential energy (so 6.9 J) E k = ½mv 2 = 5.04 J, so PE = 6.887 – 5.04 = 1.85 J E k = ½mω 2 (x o 2 - x 2 ) = ½(1.12 kg)(7.306 rad/sec) 2 ( (0.48 m) 2 – (0.23 m) 2 ) = 5.306 J ≈ 5.3 J Assuming at t = 0 it was passing through equilibrium headed in the positive direction: x = (0.48 m)sin((7.306 rad/sec)t) v = (0.48 m)(7.306 rad/sec)cos((7.306 rad/sec)t) because v o = ωx o

6. Whiteboards: Energy 11 | 2 | 3 | 4 | 523 TOC

7. An SHO has a mass of 0.259 kg, an amplitude of 0.128 m and an angular velocity of 14.7 rad/sec. What is its total energy? (save this value in your calculator) Use E T = 1 / 2 m  2 x o 2 0.458 J W

8. An SHO has a mass of 0.259 kg, an amplitude of 0.128 m and an angular velocity of 14.7 rad/sec. What is its kinetic energy when it is 0.096 m from equilibrium? What is its potential energy? Use E k = 1 / 2 m  2 ( x o 2 – x 2 ) 0.20 J, 0.26 J W

9. An SHO has a total energy of 2.18 J, a mass of 0.126 kg, and a period of 0.175 s. a) What is its maximum velocity? b) What is its amplitude of motion? Use E k = 1 / 2 mv 2 Then  = 2  /T Use E k (max) = 1 / 2 m  2 x o 2 5.88 m/s J, 0.164 m W

10. An SHO a maximum velocity of 3.47 m/s, and a mass of 0.395 kg, and an amplitude of 0.805 m. What is its potential energy when it is 0.215 m from equilibrium?  = 2  /T Use E k = 1 / 2 mv 2 Use E k (max) = 1 / 2 m  2 x o 2 Then Use E k = 1 / 2 m  2 ( x o 2 – x 2 ) Subtract kinetic from max 0.170 J W

11. A 1250 kg car bounces up and down with the following equation of motion: (in m) x = 0.170sin(4.42t) a) what is its total energy? b) what is its kinetic energy at t = 3.50 s? Use E T = 1 / 2 m  2 x o 2 Then find x from the equation: (.04007…) Then use Use E k = 1 / 2 m  2 (x o 2 – x 2 ) 353 J, 333 J W