Two springs of force constants $k_1$ and $k_2$ are connected in series. What is the effective force constant $k_{\text{eff}}$ of the combination?

Solution

1. For springs connected in series, the total extension $x$ is the sum of the individual extensions $x_1$ and $x_2$, such that $x=x_1+x_2$.
2. Using Hooke's Law $F=kx$, we substitute $x=F/k_{\text{eff}}$, $x_1=F/k_1$, and $x_2=F/k_2$ into the equation.
3. This yields the relationship $\frac{F}{k_{\text{eff}}}=\frac{F}{k_1}+\frac{F}{k_2}$, which simplifies to $\frac{1}{k_{\text{eff}}}=\frac{1}{k_1}+\frac{1}{k_2}$.
4. Solving for $k_{\text{eff}}$ gives $k_{\text{eff}}=\frac{k_1{k}_2}{k_1+k_2}$.
5. Hence the answer is (B).