Chetan Mandayam Nayakar
Last Activity: 14 Years ago
consider the behaviour of the wire to be similar to that of a spring of spring constant 'k'. at equilibrium, kl =Mg,
k=Mg/l, elastic potential energy = (k/2)l2 =Mgl/2.
when the unextended wire is released, it is stretched by 2l before it starts moving upward and exhibits damped simple harmonic motion. the lost energy ends up as thermal energy of the wire, the amplitude of SHM keeps decreasing till static equilibrium at an extension of 'l'
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