# Puzzle: Oscillating Rope and a Spring!

Here's a puzzle one of my friends in physics class gave me -

Consider a uniform rope suspended from the ceiling with the help of a spring. At what distance from the lower end (fraction of total length of the rope), the rope can be cut so that the portion hanging from the spring oscillates remaining straight?

Initially, I thought it doesn't matter from where the rope is cut - for there will always be a restoring force upwards. This restoring force would be responsible for the oscillations. However, to my surprise, he said the answer to the puzzle is to cut less than half the rope's length, so that the remaining part oscillates remaining straight.

What's going on here? Could someone help me out? I'm not able to understand why there should be a restriction on how much length is cut, for there'll always be a restoring force upwards after the portion is cut. Is something else going on here, probably because the rope is attached to a spring?

• I'm confused by what you mean when you say "suspended with the help of a spring" and "oscillate remaining straight". Apr 15, 2018 at 4:06

When you cut a tiny bit off, it’s suddenly out of equilibrium: the weight has decreased by $\lambda \Delta L g$ so the spring is stretched by $\lambda \Delta L g/k$ too much. So the system will oscillate with that amplitude and frequency $\sqrt{k \over{(L - \Delta L) \lambda}}$ and peak acceleration $A\omega^2 = (\lambda \Delta L g/k) {k\over{(L - \Delta L) \lambda}} = {\Delta L \over {L - \Delta L}}g$
Eventually, that peak acceleration will be more than $g$. That’s not a problem on the way up; the spring just pulls hard. But it is a problem on the way down, because you can’t push on a string: it can’t fall fast enough to keep tension in it.
That happens when the acceleration is $g$ or more: $\Delta L \ge L/2$