Question

Figure (a) attached 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 Fig. (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 Fig (a) and the two masses in Fig (b) are released, what is the period of oscillation in each case ?

Answer 1

(a) The maximum extension of the spring in both cases will $=F/k$, where k is the spring constant of the springs used.

(b) In Figure (a) if $x$ is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass is $F=-kx$ i.e., $F\propto x$

As this $F$ is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM.

Spring factor $=$ spring constant $=k$

inertia factor $=$ mass of the given mass $=m$

As time period,

$T=2\pi \sqrt{\frac{inertiafactor}{springfactor}}$

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

In Figure (b), we have a two body system of spring constant k and reduced mass, $\mu =m\times m/m+m=m/2$

Inertia factor $=m/2$

Spring factor $=k$

$\therefore$ time peroid, $T=2\pi \sqrt{\frac{m/2}{k}}=2\pi \sqrt{\frac{m}{2k}}$