Question

A light spring of force constant 1.0N/cm is fixed between two walls. When a force of F = 5.0 N is exerted along the spring at point C as shown. What will be the displacement of point C ?

• A. 8 mm
• B. 22.2 mm
• C. Spring's relaxed length must be known

Answer 1

Answer: (A)

8 mm

Answer 2

The spring is treated as two segments in series, but when a force is applied at the junction, the segments act in parallel in terms of restoring forces. The force constant of a spring segment is inversely proportional to its length. Let the initial lengths of the segments be $L_1$ and $L_2$, and the force constant of the entire spring be $k$. Then the force constants of the segments are $k_1 = \frac{k(L_1+L_2)}{L_1}$ and $k_2 = \frac{k(L_1+L_2)}{L_2}$. When a force F is applied at the junction, causing a displacement x, the restoring forces from the segments are $F_1 = k_1 x$ and $F_2 = k_2 x$. In equilibrium, $F = F_1 + F_2 = (k_1 + k_2) x$. The effective force constant is $k_{eff} = k_1 + k_2$. Substituting the expressions for $k_1$ and $k_2$, we get $k_{eff} = \frac{k(L_1+L_2)}{L_1} + \frac{k(L_1+L_2)}{L_2} = k(L_1+L_2) \left( \frac{1}{L_1} + \frac{1}{L_2} \right) = k(L_1+L_2) \frac{L_1+L_2}{L_1 L_2} = k \frac{(L_1+L_2)^2}{L_1 L_2}$.

So, $F = k_{eff} x = k \frac{(L_1+L_2)^2}{L_1 L_2} x$.

Given $k = 1.0$ N/cm $= 100$ N/m, $L_1 = 0.4$ m, $L_2 = 0.1$ m, $F = 5.0$ N.

$x = \frac{F}{k_{eff}} = \frac{F L_1 L_2}{k (L_1 + L_2)^2}$.

$x = \frac{5.0 \times 0.4 \times 0.1}{100 \times (0.4 + 0.1)^2} = \frac{5.0 \times 0.04}{100 \times (0.5)^2} = \frac{0.2}{100 \times 0.25} = \frac{0.2}{25} = 0.008$ m.

$x = 0.008 \text{ m} = 8 \text{ mm}$.