Deriving the Effective Spring Constant for Two Springs in Series

Problem Statement

Consider two springs arranged in series along a horizontal surface, with a wall on the left side and the springs connected to a mass on the right. Let:

• $L_1$ and $L_2$ denote the rest lengths of Spring 1 and Spring 2, respectively.
• $k_1$ and $k_2$ represent the spring constants of each spring.

The reference point (origin) is set at the right end of Spring 1 when it is at its rest length.

Define:

• $x_1$ as the position of the right end of Spring 1,
• $x_2$ as the position of the right end of Spring 2, both moving along the horizontal axis.

Since Spring 2 moves according to Spring 1’s movement, its rest position (the position of the right end of Spring 2 when it is at rest length) is given by $x_1 + L_2$.

Force Analysis

The force exerted by Spring 1 on Spring 2 is:

\[F_1 = -k_1 \Delta x_1 = -k_1 (x_1 - 0) = -k_1 x_1.\]

The force exerted by Spring 2 on the mass is:

\[F_2 = -k_2 \Delta x_2 = -k_2 (x_2 - (x_1 + L_2)) = -k_2 (x_2 - x_1 - L_2).\]

Since $F_1 = F_2$ (as the force transmitted through Spring 2 is the same as that exerted by Spring 1), we have:

\[-k_1 x_1 = -k_2 (x_2 - x_1 - L_2).\]

Expanding and rearranging terms:

\[-k_1 x_1 = k_2 x_1 - k_2 (x_2 - L_2),\]

yielding:

\[x_2 - L_2 = \frac{k_1 + k_2}{k_2} x_1.\]

Finding the Equivalent Spring Constant

We aim to find an equivalent spring constant $k_{\text{eq}}$ such that:

\[F_s = -k_{\text{eq}} (\Delta x_1 + \Delta x_2),\]

where $\Delta x_1 + \Delta x_2$ is the total displacement.

Thus,

\[F_s = -k_{\text{eq}} (x_1 + x_2 - (x_1 + L_2)) = -k_{\text{eq}} (x_2 - L_2).\]

Since $F_s$ should match $F_2$ (both representing the spring force on the mass), we set $F_s = F_2$ and find:

\[-k_{\text{eq}} (x_2 - L_2) = -k_1 x_1.\]

Rearranging gives:

\[k_{\text{eq}} (x_2 - L_2) = k_1 x_1.\]

Using our previous substitution and simplifying, we arrive at:

\[k_{\text{eq}} \frac{k_1 + k_2}{k_2} x_1 = k_1 x_1.\]

Solving for $k_{\text{eq}}$, we find:

\[k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}.\]

Alternatively, we can express this as:

\[\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}.\]

This result shows that the effective spring constant $k_{\text{eq}}$ for two springs in series is the harmonic mean of the individual spring constants.

Conclusion

We derived the equivalent spring constant for two springs in series, showing that:

\[k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2},\]

which provides insight into how forces are distributed in spring systems.