Problem Analysis & Free Body Diagram
Let $x$ be the increase in separation between the bars A and B. We denote the compression in spring $k_2$ as $x_2$ and the extension in spring $k_1$ as $x_1$.
Figure 1: Free body diagrams of Bar B and Block C showing the spring forces.
1. Determining the State of Spring $k_2$
Consider the equilibrium of Bar B. Since the bar is light (massless), the net force on it must be zero ($F_{net} = ma = 0$).
- Bar A is pulled to the right, causing the two outer springs (total stiffness $2k$) to extend by an amount $x$.
- These outer springs exert a force $2kx$ on Bar B directed to the right.
- To balance this rightward force, the inner spring $k_2$ must exert a force on Bar B directed to the left.
For spring $k_2$ (located to the right of B) to push Bar B to the left, it must be in a state of compression.
Equation for Bar B:
$$ k_2 x_2 = 2k x \implies x_2 = \frac{2k x}{k_2} \quad \text{— (i)} $$2. Geometry of Deformation
The total increase in separation between the bars ($x$) is the sum of the changes in length of the segments between them. The segment containing $k_2$ compresses by $x_2$, effectively “losing” length, while the segment containing $k_1$ extends by $x_1$, gaining length.
The vector sum of displacements yields:
$$ x = x_1 – x_2 $$Thus, the extension of spring $k_1$ is:
$$ x_1 = x + x_2 \quad \text{— (ii)} $$3. Equation of Motion for Block C
Block C accelerates to the right with acceleration $a$. The forces acting on it are:
- Force from compressed spring $k_2$ pushing to the right: $F_2 = k_2 x_2$.
- Force from extended spring $k_1$ pulling to the right: $F_1 = k_1 x_1$.
Applying Newton’s Second Law:
$$ k_1 x_1 + k_2 x_2 = ma $$Substituting $x_1$ from (ii):
$$ k_1 (x + x_2) + k_2 x_2 = ma $$ $$ k_1 x + (k_1 + k_2) x_2 = ma $$Now, substitute $x_2$ from (i):
$$ k_1 x + (k_1 + k_2) \left( \frac{2k x}{k_2} \right) = ma $$Factoring out $x$:
$$ x \left[ k_1 + \frac{2k(k_1 + k_2)}{k_2} \right] = ma $$Simplifying the term in the brackets:
$$ x \left[ \frac{k_1 k_2 + 2k k_1 + 2k k_2}{k_2} \right] = ma $$ $$ x \left[ \frac{k_1 k_2 + 2k(k_1 + k_2)}{k_2} \right] = ma $$Final Result
Solving for the separation $x$: