1 / 12

Simple Harmonic Motion - Dynamics

SOOOO:  =. Simple Harmonic Motion - Dynamics. F = ma F = -kx (Not in data packet) ma = -kx a = - kx m x = x o sin(t) v = x’ =  x o cos(t) a = x’’ = -x o  2 sin(t). Derive the energy equations: E k = 1 / 2 m  2 ( x o 2 – x 2 ) E k (max) = 1 / 2 m  2 x o 2

mai
Download Presentation

Simple Harmonic Motion - Dynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SOOOO:  = Simple Harmonic Motion - Dynamics • F = ma • F = -kx (Not in data packet) • ma = -kx • a = -kx • m • x = xosin(t) • v = x’ = xocos(t) • a = x’’ = -xo2sin(t)

  2. Derive the energy equations: Ek = 1/2m2(xo2 – x2) Ek (max) = 1/2m2xo2 ET = 1/2m2xo2 Simple Harmonic Motion - Energy Ek (max) = 1/2mvo2 Ep (max) = 1/2kxo2 Where they happen • Ek: 0 max 0 • Ep: max 0 max

  3. Simple Harmonic Motion - Energy Ek = 1/2m2(xo2 – x2) Ek (max) = 1/2m2xo2 ET = 1/2m2xo2 ET – Total Energy Ek (max) – Maximum Kinetic Energy Ek – Kinetic Energy  – “Angular” velocity T – Period of motion x – Position (at some time) v – Velocity (at some time) xo – Max Position (Amplitude) vo – Max Velocity

  4. Energy Example – An SHO has an amplitude of 0.48 m, a mass of 1.12 kg, and a period of 0.86 seconds. a) what is its angular velocity? b) what is its maximum kinetic energy? c) what is its total energy? 7.306 rad/s, 6.887 J, 6.887 J

  5. Energy Example – An SHO has an amplitude of 0.48 m, a mass of 1.12 kg, and a period of 0.86 seconds. d) what is the kinetic energy when it is 0.23 m from equilibrium? What is its potential energy here? 5.306 J, 1.581 J

  6. Energy Example – An SHO has an amplitude of 0.48 m, a mass of 1.12 kg, and a period of 0.86 seconds. e) write possible equations for its position and velocity 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 vo = ωxo

  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 ET = 1/2m2xo2 0.458 J

  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 Ek = 1/2m2(xo2 – x2) 0.20 J, 0.26 J

  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 Ek = 1/2mv2 Then  = 2/T Use Ek (max) = 1/2m2xo2 5.88 m/s J, 0.164 m

  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 Ek = 1/2mv2 Use Ek (max) = 1/2m2xo2 Then Use Ek = 1/2m2(xo2 – x2) Subtract kinetic from max 0.170 J

  11. A 1250 kg car moves 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 ET = 1/2m2xo2 Then find x from the equation: (.04007…) Then use Use Ek = 1/2m2(xo2 – x2) 353 J, 333 J

More Related