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(D) The motion is Simple Harmonic Motion $(SHM)$ between $y = 0 \ cm$ and $y = 10 \ cm$. The mean position is at $y = 5 \ cm$.The amplitude of the osc

Vedclass · Accepted Answer

$F = \frac{8}{25}(5 - y)$

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(D) The motion is Simple Harmonic Motion $(SHM)$ between $y = 0 \ cm$ and $y = 10 \ cm$. The mean position is at $y = 5 \ cm$.The amplitude of the oscillation is $A = 5 \ cm = 0.05 \ m$.From the graph,the maximum kinetic energy is $KE_{max} = 4 \ J$ at the mean position $(y = 5 \ cm)$.The formula for maximum kinetic energy in $SHM$ is $KE_{max} = \frac{1}{2} k A^2$,where $k = m\omega^2$.Substituting the values: $4 = \frac{1}{2} k (0.05)^2$.$4 = \frac{1}{2} k (0.0025) \implies k = \frac{8}{0.0025} = 3200 \ N/m$.The net force in $SHM$ is given by $F = -k(y - y_{mean})$,where $y_{mean} = 5 \ cm = 0.05 \ m$.$F = -3200(y - 0.05) = -3200y + 160$.However,looking at the options provided in terms of $cm$,we use $k_{eff} = \frac{2 KE_{max}}{A^2} = \frac{2 	imes 4}{5^2} = \frac{8}{25} \ N/cm$.Thus,the net force is $F = -\frac{8}{25}(y - 5) = \frac{8}{25}(5 - y)$.

$A$ mass on a vertical spring begins its motion at rest at $y = 0 \ cm$. It reaches a maximum height of $y = 10 \ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy $(KE)$ versus position is given below. Net force on the mass varies with $y$ as

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