Continuity of tension in falling objects Imagine I'm holding a block of mass $m_1$. At the bottom of this block is a rope that is fastened to another block of mass $m_2$. We're in a uniform gravitational field $g$. In a minute I'm going to let go of the top block. Neglect air resistance, and assume I can let go instantly.
While I'm holding the top block, there is tension in the rope. When I let go, both blocks accelerate uniformly under gravity, so there's no tension in the rope.
My question is, how does the tension move between these boundary conditions? And how is this behavior different for unstretchable vs. stretchable ropes?
 A: Suppose you start by considering the two masses joined by a spring with some force constant $k$. Initially the string is stretched by the weight of the lower mass, and the extension of the spring is the usual Hooke's law expression:
$$\Delta x = \frac{F}{k} = \frac{gm_2}{k}$$
When you release the top mass the system of the two springs is in free fall so it behaves like the same system floating in space. The system oscillates sinusoidally with a frequency given by:
$$ f = \frac{1}{2\pi}\sqrt{\frac{k}{M}} $$
where $M$ is the reduced mass. So the tension in the spring will oscillate sinusoidally as well.
Now, your unstretchable rope is the limit of $k \rightarrow \infty$, so we just need to consider what happens in this limit. As we increase $k$ the initial extension $\Delta x$ decreases and the frequency increases until in the limit of $k \rightarrow \infty$ we have an oscillation with infinite frequency but zero amplitude.
This is clearly unphysical, but then the notion of a completely unstretchable rope is also unphysical as all materials have some compliance. In any real system the tension will vary smoothly and sinusoidally so there is no discontinuity in the tension. In a real rope the motion will be heavily damped so the amplitude of the oscillation will decay quickly to effectively zero.
Response to comment:
With an elastic rope rather than a spring the rope starts out stretched so on release the masses accelerate towards each other. However as soon as the extension falls to zero the rope goes slack and ceases to apply any force to the masses. So the masses will continue to move towards each other at constant velocity and eventually collide.
To calculate this final velocity start with the equation for the initial displacement:
$$\Delta x = \frac{gm}{k}$$
I'm going to assume the two masses are equal for convenience - making the masses different doesn't change the behaviour but just makes the working more complicated.
Anyhow the elastic potential energy stored in the spring is $\tfrac{1}{2}k(\Delta x)^2$, and when the rope becomes slack all the potential energy will be converted into the kinetic energy of the two masses. So in the centre of mass frame we get:
$$ 2 \times \tfrac{1}{2}mv^2 = \tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}k \left(\frac{gm}{k}\right)^2 $$
And rearranging this gives us the velocity of the masses when the rope goes slack:
$$ v = \sqrt{\frac{g^2m}{2k}} $$
If we again approach the inextensible rope by letting $k \rightarrow \infty$ we find that $v \rightarrow 0$. That is if the two masses are joined by a perfectly inextensible rope then their separation doesn't change when you release them.
A: Suppose that the rope was perfectly rigid, meaning that it reacts instantly to any forces, and the forces on one end are instantaneously transferred to the other end. Then, yes, you do get a discontinuity in tension. This is the price you pay for making approximations.
For a stretchable rope, things get a bit more complicated. Ropes typically aren't all that stretchable, but they're pretty strong, so it seems that they would respond with a restoring force following Hooke's Law, $F=kx$, for spring coefficient $k$ and displacement $x$. In equilibrium, the spring force required to hold up the bottom block keeps the rope stretched a certain amount. Once the rope is let go, the system (if you follow its center of mass as it falls) essentially looks like a spring with a mass on each end that is stretched and then let go. Therefore, the tension oscillates sinusoidally. Over time, various dissipative forces within the rope weaken the oscillations, so that eventually the rope is once again slack.
