Inside a lift, a spring (force constant k = 1 | vedclass

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(B) Initially, the spring is compressed by $x_0 = \frac{mg}{k}$ due to the weight of the block.When the lift accelerates upwards with $a = g$, the eff

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$3$

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(B) Initially, the spring is compressed by $x_0 = \frac{mg}{k}$ due to the weight of the block.When the lift accelerates upwards with $a = g$, the effective acceleration due to gravity becomes $g_{eff} = g + a = 2g$.The new equilibrium position (mean position) of the block in the frame of the lift is at a compression $x_{eq} = \frac{mg_{eff}}{k} = \frac{m(2g)}{k} = \frac{2mg}{k}$.The block starts from the initial position $x_0 = \frac{mg}{k}$ and moves to the new equilibrium position $x_{eq} = \frac{2mg}{k}$.Since the block starts from rest at $x_0$, the new equilibrium position is the mean position of the oscillation, and the initial position is an extreme position.The amplitude of oscillation is $A = x_{eq} - x_0 = \frac{2mg}{k} - \frac{mg}{k} = \frac{mg}{k}$.The maximum compression occurs at the other extreme position, which is at a distance $A$ below the new equilibrium position.Therefore, $x_{max} = x_{eq} + A = \frac{2mg}{k} + \frac{mg}{k} = \frac{3mg}{k}$.Substituting the values: $m = 1 \ kg$, $g = 10 \ m/s^2$, $k = 1000 \ N/m$.$x_{max} = \frac{3 	imes 1 	imes 10}{1000} = \frac{30}{1000} \ m = 0.03 \ m = 3 \ cm$.

Inside a lift, a spring (force constant $k = 1000 \ N/m$) and a block $(mass = 1 \ kg)$ are both in a state of rest. Now, the lift suddenly starts moving upwards with an acceleration $a = g$. Find the maximum total compression in the spring in centimeters. $(g = 10 \ m/s^2)$

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