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UPSC 2026 Maths Optional Paper 1 Q7a — Step-by-Step Solution

15 marks · Section B

Simple harmonic motion (free, damped, forced) · Dynamics & Statics · asked 9× in 14 yrs · Read the full method →

Question

A light elastic string of natural length $l$ has one extremity fixed at a point $A$ and the other attached to a stone (mass $m$) the weight of which in equilibrium would extend the string to a length $b$. Show that if the stone be dropped from rest at $A$, it will come to instantaneous rest at a depth $\sqrt{b^2 - l^2}$ below the equilibrium position and this depth is attained in time

Prove also that if the greatest depth below $A$ be $l\cot^2\dfrac{\theta}{2}$, then the modulus of elasticity is $\dfrac{1}{2}mg\tan^2\theta$.

Technique

Two-phase rectilinear motion: free fall while the string is slack (fall of $l$), then SHM about the equilibrium position once the string is taut, with $\omega^2 = \lambda/(ml) = g/(b-l)$. Get the amplitude from the energy relation $v^2 = \omega^2(a^2 - x^2)$ at the junction instant, and the phase time by extending the SHM back to its (fictitious) upper extreme.

Solution

Setting up: geometry, Hooke’s law, and the two phases

Take the vertical line through $A$, measuring depth downwards from $A$.

Modulus from the equilibrium condition. In equilibrium the string has length $b$, i.e. extension $b - l$. Hooke’s law gives the tension $T = \lambda\,\dfrac{\text{extension}}{\text{natural length}}$, and $T$ balances the weight:

The equilibrium position $E$ is therefore at depth $b$ below $A$.

Phase 1 — free fall (string slack). The string is light and elastic: it exerts no force until it is stretched beyond its natural length. So while the stone is between depth $0$ and depth $l$, the only force is gravity.

Phase 2 — taut string (SHM). Beyond depth $l$ the string pulls up with tension $\lambda(\text{ext})/l$. Let $x$ be the displacement of the stone below $E$, so depth below $A$ is $b + x$ and the extension is $b + x - l$. Then

Hence

So Phase 2 is simple harmonic about $E$ with $\omega = \sqrt{g/(b-l)}$.

Phase 1: the junction data

Falling freely from rest through a distance $l$:

At this instant the stone is at depth $l$, i.e. at

Amplitude of the SHM — the depth of instantaneous rest

For SHM of amplitude $a$ about $E$, $\;v^2 = \omega^2\left(a^2 - x^2\right)$. Substituting the junction data $(x_1, v_1)$ and $\omega^2$ from (2):

The stone comes to instantaneous rest at the lower extreme of this SHM, i.e. at a depth $a = \sqrt{b^2-l^2}$ below the equilibrium position, as required.

Consistency: $a = \sqrt{(b-l)(b+l)} > b-l$ since $b+l > b-l$, so the lower extreme lies genuinely below $E$ and the string stays taut (indeed stretches further) throughout Phase 2 — the SHM law (2) is valid over the whole of Phase 2. At that lowest point the restoring force $-\lambda a/l \ne 0$, so the rest is instantaneous, not an equilibrium.

Time to reach that depth

Write the Phase-2 SHM as

where $\tau$ is measured from the (fictitious) instant at which the stone would have been at the upper extreme $x = -a$. This is legitimate: any SHM about $E$ with amplitude $a$ has this form for a suitable time origin, and we only use it on the interval actually traversed. With this choice $x$ increases (stone descends) for $0 < \omega\tau < \pi$, and the lower extreme $x = +a$ is reached at $\omega\tau = \pi$.

Let $\tau_1$ be the value of $\tau$ at the junction, where $x = -(b-l)$:

Hence the time spent in Phase 2 is

using $1/\omega = \sqrt{(b-l)/g}$ from (2). Adding (3):

The modulus in terms of $\theta$

The greatest depth below $A$ is (depth of $E$) $+$ (amplitude):

Invert (4) for $b$. From $\sqrt{b^2-l^2} = D - b$ (note $D > b$), squaring:

and therefore the extension in equilibrium is

Substituting into (1):

Put $D = l\cot^2\dfrac{\theta}{2}$. Write $k = \cot^2\dfrac{\theta}{2}$, so $D = lk$ and $D - l = l(k-1)$. Then (5) becomes

Now use the double-angle identity $\cot\theta = \dfrac{\cot^2\frac{\theta}{2} - 1}{2\cot\frac{\theta}{2}}$, i.e.

Hence

(Since $D = b + \sqrt{b^2-l^2} > l$ we have $k > 1$, i.e. $\cot\frac{\theta}{2} > 1$, i.e. $0 < \theta < \frac{\pi}{2}$ — $\theta$ is acute, so $\cot\theta > 0$ and the square root taken above has the right sign.)

Answer