ch15-p060 - 60(a From Hookes law we have k= 500 kg 9.8 m/s...

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() 2 2 500 kg 9.8 m/s = 4.9 10 N/cm. 10cm k (b) The amplitude decreasing by 50% during one period of the motion implies eT bT m == 2 1 2 2 where π ω . Since the problem asks us to estimate, we let ≈= km / . That is, we let ≈≈ 49000 500 , N/m kg 9.9 rad / s so that T 0.63 s. Taking the (natural) log of both sides of the above equation, and rearranging, we find 3 2 500 kg 2 ln2 0.69 1.1 10 kg/s. 0.63 s m b T =≈ = × Note: if one worries about the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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