12_03_SHM_4

# 12_03_SHM_4 - Now find all the velocities at each location...

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12/3 Simple Harmonic Motion 2 Presentation “Energy, Vetocity, and Accel. in SHM” 2 HW “Simple Harmonic Motion 2” Due Wednesday 12/5 2 Exam IV Thursday Eve 12/6 5-7 Wit 116 Angular Momentum Simple Harmonic Motion Torque Energy

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Key Concepts: The period T depends on k and m only. The spring constant can be found from Hooke’s law or the period. The acceleration is found with F net = ma. The velocity is found with energy. Can’t use v ave , accel. not constant
Apply Energy to the SHM system (frictionless) and fill the energy buckets at each location. m k KE PE S KE PE S KE PE S KE PE S KE PE S m = 2kg, k = 600N/m, A = 0.2m 0 = 1 / 2 k x 2 = 12J 0 = 12J = 12J = 12J = 12J Energy is conserved, no external forces. 0 0 ? ? PE S = 1 / 2 k x 2 x Is half as big so PE S is. .. PE S =3J KE =9J

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Unformatted text preview: Now find all the velocities at each location. m k KE PE S KE PE S KE PE S KE PE S KE PE S m = 2kg, k = 600N/m, A = 0.2m 12J Energy is conserved, no external forces. 3J 9J 12J 12J 12J v = 0 v = 0 v = 3.5m/s KE = 1 / 2 mv 2 v = 3.5m/s v = 3.0m/s Now find the acceleration at each location. m k m = 2kg, k = 600N/m, A = 0.2m We need to draw the FBDs. F Spring = -k ∆ x a = 60m/s 2 a = 60m/s 2 a = 30m/s 2 a = 0 a = 0 = ma T S,M T S,M T S,M ∆ x The big picture m k m = 2kg, k = 600N/m, A = 0.2m F Spring = -k ∆ x = ma T S,M T S,M T S,M a = 60m/s 2 v = a = 60m/s 2 v = a = v = 3.5m/s a = v = 3.5m/s a = 30m/s 2 v = 3.0m/s Acceleration and velocity...
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## This note was uploaded on 12/21/2010 for the course PHYS Physics 21 taught by Professor Larrywatson during the Fall '08 term at North Dakota.

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12_03_SHM_4 - Now find all the velocities at each location...

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