Simple Harmonic Motion – Dynamics and Energy Contents:

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Presentation transcript:

Simple Harmonic Motion – Dynamics and Energy Contents: Example Whiteboards

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 = xosin(t) v = x’ = xocos(t) a = x’’ = -xo2sin(t) SOOOO:  = TOC

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

Simple Harmonic Motion - Energy Ek = 1/2m2(xo2 – x2) ET = 1/2m2xo2 ET – Total 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 TOC

a) what is its angular velocity? 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? d) what is the kinetic energy when it is 0.23 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 Ek(max) = ½mω2xo2 = 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 (so 6.9 J) Ek = ½mω2(xo2 - x2) = 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 vo = ωxo TOC

Whiteboards: Energy 1 | 2 | 3 | 4 | 5 TOC

What is its total energy? (save this value in your calculator) 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 W 0.458 J

An SHO has a mass of 0. 259 kg, an amplitude of 0 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) W 0.20 J, 0.26 J

a) What is its maximum velocity? b) What is its amplitude of motion? 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 W 5.88 m/s J, 0.164 m

An SHO a maximum velocity of 3. 47 m/s, and a mass of 0 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 W 0.170 J

a) what is its total energy? 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 ET = 1/2m2xo2 Then find x from the equation: (.04007…) Then use Use Ek = 1/2m2(xo2 – x2) W 353 J, 333 J