Question

At a smooth horizontal table there are two identical cubes each of mass m, connected by a spring of spring constant k. The length of spring in unstretched state is $l_0$. The right cube is linked to a load mass m at the end as shown. At some time, the system is released from rest when spring is unstretched. Find the maximum distance (in cm) between blocks during the motion of the system. [$l_0 = 1cm, m = 3kg, k = 1000 N/m, g = 10 m/s^2$]

Answer 1

4 cm

Answer 2

Solution Explanation:
Let the extension of the spring be

$$u = (x_2 - x_1) - l_0,$$

where $l_0$ is the natural length (1 cm).
For the left cube (mass $m$):

$$m\,x_1'' = k\,u.$$

For the right cube, which is pulled by an additional force $mg$ (via the hanging mass):

$$m\,x_2'' = mg - k\,u.$$

Subtracting these, the relative acceleration is

$$u'' = x_2'' - x_1'' = \frac{mg - k\,u - k\,u}{m} = \frac{mg - 2k\,u}{m}.$$

This gives

$$u'' + \frac{2k}{m}\,u = g.$$

Define $\omega^2 = \frac{2k}{m}$. The equilibrium (steady state) extension $u_{\text{eq}}$ satisfies

$$\frac{2k}{m}\,u_{\text{eq}} = g\quad\Longrightarrow\quad u_{\text{eq}} = \frac{mg}{2k}.$$

But notice that the system is released from rest with $u(0)=0$ and $u'(0)=0$. Its solution is

$$u(t)=u_{\text{eq}}\left(1-\cos(\omega t)\right)=\frac{mg}{2k}\left(1-\cos\Bigl(\sqrt{\frac{2k}{m}}\,t\Bigr)\right).$$

The maximum value of $1-\cos(\omega t)$ is $2$ (attained when $\cos(\omega t)=-1$). Hence the maximum extension is

$$u_{\max} = \frac{mg}{2k}\times 2 = \frac{mg}{k}.$$

Substitute given values:

$$u_{\max} = \frac{3\times 10}{1000} = \frac{30}{1000}=0.03\,{\rm m}=3\,{\rm cm}.$$

Thus the maximum distance between the cubes (i.e. the spring’s length) is

$$l_0+u_{\max} = 1\,{\rm cm} + 3\,{\rm cm} = 4\,{\rm cm}.$$