← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q6b — Solution

15 marks · Section B

Question

One end of an elastic string, having natural length $a$, is fixed at some point $O$ and a heavy particle is attached to the other end of the string. The string is drawn vertically downward till it is four times its natural length at the point $C$ and then released. If the modulus of elasticity of the string is equal to the weight of the particle, then show that the particle will return to the same point $C$ in the time $\sqrt{\dfrac{a}{g}}\left(2\sqrt{3} + \dfrac{4\pi}{3}\right)$.

Technique

The motion has two phases: SHM while the string is taut (Hooke’s-law tension), and free projectile flight while the string is slack. Match position and velocity at the transition, and use symmetry to assemble the total time.

Solution

Let $y$ denote the distance of the particle below $O$ (downward positive). The string has natural length $a$ and modulus of elasticity $\lambda$. By Hooke’s law, when stretched to length $y>a$ the tension is $T = \lambda\,\frac{\text{extension}}{\text{natural length}} = \lambda\,\frac{y-a}{a}.$ Given $\lambda = mg$ (modulus equals the particle’s weight).

Equilibrium and SHM (string taut, $y \ge a$)

Newton’s law (taking downward positive), while taut: $m\ddot y = mg - T = mg - mg\frac{y-a}{a} = mg\,\frac{2a - y}{a} = -\frac{mg}{a}\,(y - 2a).$ This is SHM about the equilibrium length $y = 2a$ with angular frequency $\omega^2 = \frac{g}{a},\qquad \omega = \sqrt{\frac{g}{a}}.$

The particle is released from rest at $C$, where the string is $4a$ long, so $y = 4a$. Hence the SHM has centre $y=2a$ and amplitude $A = 4a - 2a = 2a.$

Measuring time $t$ from release, $y(t) = 2a + 2a\cos(\omega t).$

Phase 1 — taut, from $C$ up to the natural length

The string remains taut until the length returns to its natural value $a$, i.e. $y = a$: $2a + 2a\cos(\omega t) = a \;\Longrightarrow\; \cos(\omega t) = -\tfrac{1}{2} \;\Longrightarrow\; \omega t_1 = \frac{2\pi}{3}.$ So $t_1 = \frac{2\pi}{3\omega} = \frac{2\pi}{3}\sqrt{\frac{a}{g}}.$

Speed at this instant (the string just goes slack): $\dot y = -2a\omega\sin(\omega t_1) = -2a\omega\sin\frac{2\pi}{3} = -2a\omega\cdot\frac{\sqrt3}{2} = -\sqrt3\,a\omega.$ With $\omega=\sqrt{g/a}$: the speed is $v_0 = \sqrt3\,a\sqrt{g/a} = \sqrt{3ag}$ (directed upward, toward $O$).

Phase 2 — slack string, free flight

Once $y<a$ the string is slack ($T=0$); the particle is a free body under gravity alone, moving upward with speed $v_0=\sqrt{3ag}$, decelerating at $g$. It rises, stops, and falls back to the same level $y=a$ with the same speed $v_0$ (energy/symmetry). The time of this up-and-down free flight is $t_2 = \frac{2v_0}{g} = \frac{2\sqrt{3ag}}{g} = 2\sqrt{\frac{3a}{g}} = 2\sqrt3\,\sqrt{\frac{a}{g}}.$

Phase 3 — taut again, back to $C$

At $y=a$ the string becomes taut again, now moving downward with speed $v_0$. By time-reversal symmetry of the SHM, the particle retraces phase 1 in reverse and reaches $C$ ($y=4a$) after a further time $t_1$.

Total time

$T_{\text{return}} = t_1 + t_2 + t_1 = 2t_1 + t_2 = \frac{4\pi}{3}\sqrt{\frac{a}{g}} + 2\sqrt3\,\sqrt{\frac{a}{g}} = \sqrt{\frac{a}{g}}\left(2\sqrt3 + \frac{4\pi}{3}\right).$

Answer

$\boxed{\;T_{\text{return}} = \sqrt{\frac{a}{g}}\left(2\sqrt{3} + \frac{4\pi}{3}\right).\;}$