Figure $(a)$ shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. $A$ force $F$ applied at the free end stretches the spring. Figure $(b)$ shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Figure $(b)$ is stretched by the same force $F$.
$(a)$ What is the maximum extension of the spring in the two cases?
$(b)$ If the mass in Figure $(a)$ and the two masses in Figure $(b)$ are released,what is the period of oscillation in each case?

(N/A) For the one-block system (Figure $(a)$):
When a force $F$ is applied to the free end,the extension $l$ is given by $F = kl$,so $l = F/k$.
The equation of motion is $m(d^2x/dt^2) = -kx$. This is simple harmonic motion with angular frequency $\omega = \sqrt{k/m}$.
The time period is $T = 2\pi/\omega = 2\pi\sqrt{m/k}$.
For the two-block system (Figure $(b)$):
Each end is pulled by force $F$. The spring is stretched by $l$ such that the tension in the spring is $F$. Thus,$F = kl$,which gives $l = F/k$. The extension is the same in both cases.
For the oscillation,consider the center of mass of the spring. Each mass $m$ moves relative to the center of mass. The effective spring constant for each half of the spring is $k' = 2k$. The equation of motion for one mass is $m(d^2x/dt^2) = -2kx$.
This gives $\omega = \sqrt{2k/m}$.
The time period is $T = 2\pi/\omega = 2\pi\sqrt{m/(2k)}$.