Solution
We analyze the restoring force characteristics of the taut cord for small displacements and then apply the Work-Energy theorem.
1. Force Analysis (Small Deflection):
Let the depression at the midpoint be $y$. The length of each half of the cord becomes $\sqrt{(L/2)^2 + y^2}$.
Since $y \ll L$ (cm vs m), we approximate the change in length. The total extended length $L’$ is: $$ L’ = 2 \sqrt{\frac{L^2}{4} + y^2} \approx 2 \left( \frac{L}{2} + \frac{y^2}{L} \right) = L + \frac{2y^2}{L} $$ The extension $\Delta L = \frac{2y^2}{L}$.
The tension in the cord is $T = k \Delta L = \frac{2ky^2}{L}$.
The restoring vertical force $F_y$ is the component of tension from both sides: $$ F_y = 2T \sin\theta \approx 2T \tan\theta = 2 \left( \frac{2ky^2}{L} \right) \left( \frac{y}{L/2} \right) $$ $$ F_y = \frac{4ky^2}{L} \cdot \frac{2y}{L} = \frac{8k}{L^2} y^3 $$ Notice that the restoring force is proportional to $y^3$. Let $C = \frac{8k}{L^2}$, so $F_y = C y^3$.
Let the depression at the midpoint be $y$. The length of each half of the cord becomes $\sqrt{(L/2)^2 + y^2}$.
Since $y \ll L$ (cm vs m), we approximate the change in length. The total extended length $L’$ is: $$ L’ = 2 \sqrt{\frac{L^2}{4} + y^2} \approx 2 \left( \frac{L}{2} + \frac{y^2}{L} \right) = L + \frac{2y^2}{L} $$ The extension $\Delta L = \frac{2y^2}{L}$.
The tension in the cord is $T = k \Delta L = \frac{2ky^2}{L}$.
The restoring vertical force $F_y$ is the component of tension from both sides: $$ F_y = 2T \sin\theta \approx 2T \tan\theta = 2 \left( \frac{2ky^2}{L} \right) \left( \frac{y}{L/2} \right) $$ $$ F_y = \frac{4ky^2}{L} \cdot \frac{2y}{L} = \frac{8k}{L^2} y^3 $$ Notice that the restoring force is proportional to $y^3$. Let $C = \frac{8k}{L^2}$, so $F_y = C y^3$.
2. Case 1: Static Equilibrium:
The load descends gradually to $y_1 = 1 \text{ cm}$. At this point, the restoring force balances the weight $mg$. $$ mg = C y_1^3 \quad \text{— (i)} $$
The load descends gradually to $y_1 = 1 \text{ cm}$. At this point, the restoring force balances the weight $mg$. $$ mg = C y_1^3 \quad \text{— (i)} $$
3. Case 2: Dynamic Drop:
The load is dropped from a height $h$ above the cord. It sticks and descends to a maximum depth $y_2 = 2 \text{ cm}$.
Using the Work-Energy Theorem between the drop point and the lowest point: $$ W_{gravity} + W_{spring} = \Delta K = 0 $$ The mass falls a total distance $h + y_2$. The work done by the non-linear spring force is $\int F_y dy$. $$ mg(h + y_2) – \int_{0}^{y_2} (C y^3) dy = 0 $$ $$ mg(h + y_2) = \left[ \frac{C y^4}{4} \right]_0^{y_2} = \frac{C y_2^4}{4} $$ Substitute $mg = C y_1^3$ from equation (i): $$ C y_1^3 (h + y_2) = \frac{C y_2^4}{4} $$ $$ h + y_2 = \frac{1}{4} \frac{y_2^4}{y_1^3} $$ $$ h = \frac{1}{4} y_2 \left( \frac{y_2}{y_1} \right)^3 – y_2 $$
The load is dropped from a height $h$ above the cord. It sticks and descends to a maximum depth $y_2 = 2 \text{ cm}$.
Using the Work-Energy Theorem between the drop point and the lowest point: $$ W_{gravity} + W_{spring} = \Delta K = 0 $$ The mass falls a total distance $h + y_2$. The work done by the non-linear spring force is $\int F_y dy$. $$ mg(h + y_2) – \int_{0}^{y_2} (C y^3) dy = 0 $$ $$ mg(h + y_2) = \left[ \frac{C y^4}{4} \right]_0^{y_2} = \frac{C y_2^4}{4} $$ Substitute $mg = C y_1^3$ from equation (i): $$ C y_1^3 (h + y_2) = \frac{C y_2^4}{4} $$ $$ h + y_2 = \frac{1}{4} \frac{y_2^4}{y_1^3} $$ $$ h = \frac{1}{4} y_2 \left( \frac{y_2}{y_1} \right)^3 – y_2 $$
4. Calculation:
Given $y_1 = 1 \text{ cm}$ and $y_2 = 2 \text{ cm}$. $$ \frac{y_2}{y_1} = 2 $$ $$ h = \frac{1}{4} (2) (2)^3 – 2 $$ $$ h = \frac{1}{4} (2) (8) – 2 $$ $$ h = 4 – 2 = 2 \text{ cm} $$ The height above the cord from where the load was dropped is **2 cm**.
Given $y_1 = 1 \text{ cm}$ and $y_2 = 2 \text{ cm}$. $$ \frac{y_2}{y_1} = 2 $$ $$ h = \frac{1}{4} (2) (2)^3 – 2 $$ $$ h = \frac{1}{4} (2) (8) – 2 $$ $$ h = 4 – 2 = 2 \text{ cm} $$ The height above the cord from where the load was dropped is **2 cm**.