Correct Option: (b)

Let’s analyze the transition from state 1 (Gravity ON) to state 2 (Gravity OFF).

Initial State (Gravity ON):

• The equilibrium position is stretched by $\delta = mg/k$ from the natural length. Let’s define the natural length position as $x=0$. So, Old Mean $= mg/k$.
• The block oscillates with amplitude $A_{old}$.
• The “lowest position” corresponds to $x = (mg/k) + A_{old}$. At this point, the velocity of the block is zero ($v=0$).

The Switch (Gravity OFF):

• At the exact moment the block is at its lowest position, gravity vanishes.
• The only force acting on the block is the spring force $F_s = -kx$.
• The new equilibrium position is the natural length of the spring ($x=0$), because without gravity, there is no weight to stretch the spring at rest.

New Oscillation Parameters:

• Position: The block is currently at $x = (mg/k) + A_{old}$.
• Velocity: The velocity is still $0$.
• Since the velocity is zero at this position relative to the new equilibrium ($x=0$), this position represents the new extreme point (Amplitude).
• New Amplitude ($A_{new}$): The distance from the new mean ($0$) to the current position. $$ A_{new} = \frac{mg}{k} + A_{old} $$ This implies $A_{new} > A_{old}$. The amplitude increases.

Graph Interpretation: The graph should show the oscillation shifting its mean position “upwards” (to the natural length line) and the amplitude becoming significantly larger. Graph (b) correctly depicts this increase in amplitude and shift in mean.