qz10sm08sol - August6,2008 PHY2053Discussion...

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August 6, 2008  PHY2053 Discussion Quiz 10 (Chapter 13.5-14.11) Name:                                   UFID: **1. (5pts) A simple pendulum with a length of 1.50 m is suspended from the ceiling of  a train. What is the period of simple harmonic motion for the pendulum if the train is  moving up an incline of 15.0˚ with an acceleration of 3.00 m/s²? The pendulum is said to be in equilibrium if the pendulum is stationary relative to the   train. We calculate the tension of the pendulum in equilibrium, and equate it with mg eff Taking +x axis along the incline and up and +y perpendicular to the incline, we get x: ma = Fsin -mgsin15˚  θ  F²sin²  = m²(a+gsin15˚)² θ y: 0 = Fcos -mgcos15˚  θ  F²cos²  = m²cos²15˚ θ  F = m√((a+gsin15˚)²+ cos²15˚) = mg eff where g eff  = √((a+gsin15˚)²+ cos²15˚) = √{(3+9.8sin15˚)²+cos²15˚} = 11.0 m/s Thus the period of the pendulum is T = 2 π √(L/g eff ) = 2×3.14×√(1.5/11) = 2.32 s
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This note was uploaded on 12/13/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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qz10sm08sol - August6,2008 PHY2053Discussion...

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