Two springs of force constants ${\mathrm{k}}_1 \mathrm{and} {\mathrm{k}}_2$ are connected to a mass m placed on a horizontal frictionless surface as shown in Figs (a) and (b). What is the ratio of the time periods of horizontal oscillation in cases (a) and (b) if ${\mathrm{k}}_1={\mathrm{k}}_2$?

In case (a) the springs are connected in series. Therefore, the spring constant of the combination is given by
$$\begin{gathered} \frac{1}{{\mathrm{k}}^{\text{'}}}=\frac{1}{{\mathrm{k}}_1}+\frac{1}{{\mathrm{k}}_2}=\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{k}}=\frac{2}{\mathrm{k}} \\ (\left.\because {\mathrm{k}}_1={\mathrm{k}}_2=\mathrm{k}\right) \end{gathered}$$ 

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{k}}{2}$.

Therefore, for case (a),
${\mathrm{T}}_{\mathrm{a}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{k}}^{\text{'}}}}=2\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{\mathrm{k}}}$
In case (b) the spring constant of the combination is
$$\begin{gathered} {\mathrm{k}}^{\text{'}\text{'}}={\mathrm{k}}_1+{\mathrm{k}}_2=2\mathrm{k} \\ \therefore {\mathrm{T}}_{\mathrm{b}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{k}}^{\text{'}\text{'}}}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}} \end{gathered}$$
Thus $\frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{b}}}=2$