Question

A vertically kept spring has masses $m$ and $2m$ attached to it as shown in the figure. The maximum amplitude of oscillation of $m$ for which $2m$ will not lift up from the ground is:

• A. g/K

Answer 1

Answer: (A)

A = g/K

Answer 2

We start from the static equilibrium. At equilibrium:

• For mass m (which hangs freely), only its weight mg is balanced by the upward spring force. Thus, $K\delta_0 = mg \implies \delta_0 = \frac{mg}{K}$.

• For mass 2m resting on the ground, the spring pulls upward with the same force mg. The ground supplies a normal force N such that $N + mg = 2mg \implies N = mg$.

Now, if m is displaced downward by an amplitude A, the spring gets stretched by an extra amount so that its extension becomes $\delta = \delta_0 + A$. Then the spring force becomes $F_s = K(\delta_0 + A) = mg + KA$.

This extra force (KA) acts upward on mass 2m. For 2m to remain in contact with the ground its net upward force must not overcome its weight. At the limiting case of losing contact the normal force becomes zero and we have: $mg + KA = 2mg$.

Solving for A gives: $KA = mg \implies A = \frac{g}{K}$.

Thus, the maximum amplitude of oscillation of m (when it is displaced downward) that ensures mass 2m does not lose contact with the ground is $A = \frac{g}{K}$.

Explanation (minimal):

1. At equilibrium, $K(\frac{mg}{K}) = mg$.
2. Displacing m downward by A increases spring extension to $\delta_0 + A$, so spring force becomes $mg + KA$.
3. To keep 2m on the ground, require $mg + KA \le 2mg$; equality gives $A = \frac{g}{K}$.