1. System Analysis

Consider a uniform rope of mass $M$ and length $L$. The mass per unit length is $\lambda = \frac{M}{L}$.
Before the cut, the system is in equilibrium. The spring force $F_s$ balances the total weight of the rope: $$F_s = Mg = (\lambda L)g$$

2. Dynamics Immediately After the Cut

The rope is cut at a height $y$ from the lower end.

• Mass of removed portion: $m_{cut} = \lambda y$
• Mass of remaining portion: $m_{rem} = \lambda (L – y)$

Immediately after the cut, the extension in the spring has not changed, so the upward force remains $F_s = \lambda L g$. However, the downward gravitational force is now reduced to the weight of the remaining portion $W’ = m_{rem}g$.

Using Newton’s Second Law for the remaining portion (taking upward as positive): $$ F_{net} = F_s – W’ $$ $$ F_{net} = (\lambda L g) – \lambda(L-y)g $$ $$ F_{net} = \lambda g [ L – (L-y) ] = \lambda y g $$

The initial upward acceleration $a$ is: $$ a = \frac{F_{net}}{m_{rem}} = \frac{\lambda y g}{\lambda(L-y)} = \left( \frac{y}{L-y} \right)g $$ This acceleration represents the maximum acceleration amplitude ($a_{max}$) of the resulting Simple Harmonic Motion (SHM).

3. Condition for the Rope to Remain Straight

For a flexible rope to oscillate without “buckling” or going slack, the internal tension must remain non-negative throughout the motion. The most critical point is at the top of the oscillation where the acceleration is directed downwards with magnitude $a_{max}$.

The condition for the rope to remain taut is that the effective gravity ($g_{eff}$) must point downwards: $$ g_{eff} = g – a_{max} \ge 0 $$ $$ a_{max} \le g $$

Substituting the value of $a_{max}$: $$ \left( \frac{y}{L-y} \right)g \le g $$ $$ \frac{y}{L-y} \le 1 $$ $$ y \le L – y $$ $$ 2y \le L $$ $$ y \le \frac{L}{2} $$