11_30_SHM_3 - 11/30 Simple Harmonic Motion / Presentation...

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Unformatted text preview: 11/30 Simple Harmonic Motion / Presentation Energy in the vertical SHM / HW Simple Harmonic Motion 2 Due Wednesday 12/5 / Exam IV Thursday Eve 12/6 5-7 Wit 116 Angular Momentum Simple Harmonic Motion Torque Energy Problem: A 2.0 kg block hangs from a spring and oscillates with a period T=1 s. What mass of block, hung from the same spring, would have T=2 s? 2 = T m k 2 4 Therefore, m = 8.0 kg. A 2.0 kg block oscillates on a horizontal frictionless surface with a period T = 1 s and an amplitude 10 cm. Find the spring constant k. Find the maximum acceleration of the block. Problem: (k = 79 N/m) T = 2 m k F = ma Energy in SHM A PE = KE = PE 1/2 kA 2 1/2 mv 2 KE = PE = KE = = 1/2 kA 2 0 0 0 KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position KE + PE s = E Equilibrium position Energy in SHM A PE = KE = PE 1/2 kA 2 1/2 mv 2 KE = PE = KE = = 1/2 kA 2 0 0 0 Total E does not change!...
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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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11_30_SHM_3 - 11/30 Simple Harmonic Motion / Presentation...

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