NLM O14

Step 1: Determine Stiffness of Spring Segments

Let the original spring have a natural length $L_0$ and a force constant $k_0$. The product of stiffness and length is constant for a uniform spring: $$k \cdot L = \text{constant} = k_0 L_0$$ The bead divides the spring into two parts:

• Upper Part: Length $l$. Stiffness $k_1 = \frac{k_0 L_0}{l}$
• Lower Part: Length $(L_0 – l)$. Stiffness $k_2 = \frac{k_0 L_0}{L_0 – l}$

Step 2: Analyze Forces in Equilibrium

When the bead moves down by a distance $y$:

• The upper spring stretches by $y$, exerting an upward force $F_1 = k_1 y$.
• The lower spring compresses by $y$, exerting an upward force $F_2 = k_2 y$.

For equilibrium, the total upward restoring force balances the weight $mg$: $$(k_1 + k_2) y = mg$$

Step 3: Derive the Function $y(l)$

Substitute the expressions for $k_1$ and $k_2$: $$ y = \frac{mg}{k_1 + k_2} = \frac{mg}{\frac{k_0 L_0}{l} + \frac{k_0 L_0}{L_0 – l}} $$ Simplify the denominator: $$ k_1 + k_2 = k_0 L_0 \left( \frac{1}{l} + \frac{1}{L_0 – l} \right) = k_0 L_0 \left( \frac{L_0 – l + l}{l(L_0 – l)} \right) = \frac{k_0 L_0^2}{l(L_0 – l)} $$ Now, substitute this back into the equation for $y$: $$ y = \frac{mg}{\frac{k_0 L_0^2}{l(L_0 – l)}} = \left( \frac{mg}{k_0 L_0^2} \right) \cdot l(L_0 – l) $$ $$ y = C \cdot (l L_0 – l^2) $$ where $C$ is a constant. This equation represents a downward-opening parabola with roots at $l = 0$ and $l = L_0$.