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All spring string and pulley shown in figure are light initially when both the spring were in their natural lenght the system was released from rest the maximum displace ment of the block is x×5mg/k calculate x
x1 be the displacement from spring attached to centre of pulley and x2 be the displacement from spring 2. Net displacement of mass m will be 2x1+x2.Now, T=kx2----(1)and 2T=kx1. ----(2)Now, from 1 and 2,x1=2x2----(3)Now from work energy conservation theorem,(1/2)k(x1)2 +(1/2)k(x2)2= mg(2x1+ x2)Using eq.(3),x2=2mg/kx1= 4mg/kNet displacement=2x1+x2=8mg/k +2mg/k=10mg/kOn comparing we get x= 5
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