← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q5d — Solution

10 marks · Section B

Question

Given that $A$ and $B$ are two points in the same horizontal line distant $2a$ apart. $AO$ and $BO$ are two equal heavy strings tied together at $O$ and carrying their weight at $O$. If $l$ is length of each string and $d$ is depth of $O$ below $AB$, then show that the parameter $c$ of this catenary, in which the strings hang, is given by $l^2 - d^2 = 2c^2\left[\cosh\left(\dfrac{a}{c}\right) - 1\right]$.

Technique

Use the common catenary $y = c\cosh(x/c)$ with parameter $c=H/w$ (horizontal tension over weight per unit length), apply the standard arc-length and sag relations, then eliminate using the identity $\cosh^2-\sinh^2=1$.

Solution

A uniform heavy string hangs in a catenary $y = c\cosh\!\left(\dfrac{x}{c}\right)$, with the lowest point at $x=0$, $y=c$, and parameter $c$.

By symmetry the two equal strings $AO$, $BO$ form a single symmetric catenary whose lowest point is $O$ (the junction carrying the load lies at the vertex). Take the origin of the catenary at its vertex, so the lowest point is at $x=0$, height $c$.

The supports $A,B$ are symmetric about the vertical through $O$ at horizontal distance $a$ on each side, i.e. at $x=\pm a$.

Standard catenary relations (vertex at $x=0$):

1. Arc length from the vertex ($x=0$) to the support ($x=a$): $s = c\sinh\!\left(\frac{a}{c}\right).$ Since each string runs from $O$ (the vertex) to a support, the length of each string is l = c\sinh\!\left(\frac{a}{c}\right). \tag{1}

2. Vertical rise (sag) from the vertex to the support level: $\text{height at }x=a \text{ minus height at }x=0 = c\cosh\!\left(\frac{a}{c}\right) - c = c\!\left[\cosh\!\left(\frac{a}{c}\right)-1\right].$ Since $AB$ is the support level and $O$ is the vertex, the depth of $O$ below $AB$ is d = c\!\left[\cosh\!\left(\frac{a}{c}\right) - 1\right]. \tag{2}

Eliminate $a/c$. Write $C \equiv \cosh\!\left(\dfrac{a}{c}\right)$. From (1) and the identity $\sinh^2 = \cosh^2 - 1$, $l^2 = c^2\sinh^2\!\left(\frac{a}{c}\right) = c^2\bigl(C^2 - 1\bigr) = c^2(C-1)(C+1).$ From (2), $d^2 = c^2(C - 1)^2.$ Subtract: $l^2 - d^2 = c^2(C-1)\bigl[(C+1) - (C-1)\bigr] = c^2(C-1)\cdot 2 = 2c^2\bigl[\cosh\!\left(\tfrac{a}{c}\right) - 1\bigr].$

Answer

$\boxed{\;l^2 - d^2 = 2c^2\!\left[\cosh\!\left(\frac{a}{c}\right) - 1\right].\;}$