To the free end of a spring hanging from a ri | vedclass

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(B) When the block is lowered slowly,the equilibrium position is reached when the spring force equals the gravitational force: $k d = m g$,which impli

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$2 d$

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(B) When the block is lowered slowly,the equilibrium position is reached when the spring force equals the gravitational force: $k d = m g$,which implies $d = \frac{m g}{k}$.When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. At the maximum extension $x$,the change in gravitational potential energy equals the elastic potential energy stored in the spring:$m g x = \frac{1}{2} k x^2$Solving for $x$ (where $x 
eq 0$):$x = \frac{2 m g}{k}$Since $d = \frac{m g}{k}$,we substitute this into the equation:$x = 2 d$.

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