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 etther 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?

For the one block system:

When a force $F$, is applied to the free end of the spring, an extension $l$, is produced. For the maximum extension, it can be written as:

$F=k l$

Where, $k$ is the spring constant

$I=\frac{F}{k}$

Hence, the maximum extension produced in the spring,

For the two block system:

The displacement ( $x$ ) produced in this case is

$x=\frac{l}{2}$

Net force, $F=+2 k x=2 k \frac{l}{2}$

$\therefore l=\frac{F}{k}$

For the one block system:

For mass ( $m$ ) of the block, force is written as:

$F=m a=m \frac{d^{2} x}{d t^{2}}$

Where, $x$ is the displacement of the block in time $t$ $\therefore m \frac{d^{2} x}{d t^{2}}=-k x$

It is negative because the direction of elastic force is opposite to the direction of displacement. $\frac{d^{2} x}{d t^{2}}=-\left(\frac{k}{m}\right) x=-\omega^{2} x$

Where, $\omega^{2}=\frac{k}{m}$

$\omega=\sqrt{\frac{k}{m}}$

Where, $\omega$ is angular frequency of the oscillation

$\therefore$ Time period of the oscillation, $T=\frac{2 \pi}{\omega}$

$=\frac{2 \pi}{\sqrt{\frac{k}{m}}}=2 \pi \sqrt{\frac{m}{k}}$

For the two block system:

$F=m \frac{d^{2} x}{d t^{2}}$

$m \frac{d^{2} x}{d t^{2}}=-2 k x$

It is negative because the direction of elastic force is opposite to the direction of displacement.

$\frac{d^{2} x}{d t^{2}}=-\left[\frac{2 k}{m}\right] x=-\omega^{2} x$

Where,

Angular frequency, $\omega=\sqrt{\frac{2 k}{m}}$

$\therefore$ Time period, $T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}}$