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

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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
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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? π 2 = T m k × 2 × 4 Therefore, m = 8.0 kg.
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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 π m k F = ma
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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
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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KE + PE s = E Equilibrium position
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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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