()22500 kg 9.8 m/s=4.9 10 N/cm.10cmk=×(b) The amplitude decreasing by 50% during one period of the motion implies eTbTm−==′2122whereπω.Since the problem asks us to estimate, we let ′≈=km/. That is, we let ′≈≈49000500,N/mkg9.9 rad / sso that T≈0.63 s. Taking the (natural) log of both sides of the above equation, and rearranging, we find 32 500 kg2ln20.691.1 10 kg/s.0.63 smbT=≈=×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.