Cross-section of a string at a point is inversely proportional to tension. Prove that the curve is a parabola.

Question

Q.) A non uniform string hangs under gravity. Its cross section at any point is inversely proportional to the tension at that point. Prove that the curve in which string hangs is an arc of the parabola with its axis vertical.

Please help me out.

Answer 1

Let $\delta m$ be an element of mass, whose length is $\delta s$, and whose cross section is $A$. Let the string have constant mass per unit volume $\rho$. Let $O$ be the origin at the lowest point on the string and let $P$ be a typical point on the string, where the arc length $OP=s$, the tension in the string at $P$ is $T$, and the tangent to the string at $P$ makes angle $\psi$ With the horizontal.

Considering the element of mass at $P$, whose length is $\delta s$, we have $$\delta m=A\rho\delta s=\frac{\rho k}{T}\delta s,$$ where $k$ is the constant of proportionality between cross section and tension. From this we have $$\frac{ds}{dm}=\frac{T}{\rho k}$$

Now consider the whole length of string $OP$ which has mass $m$, and is in equilibrium due to the horizontal tension $T_0$ at $O$ and tension $T$ at $P$.

We have $$T\cos \psi=T_0\Rightarrow T=T_0\sec \psi$$ and $$T\sin \psi=mg$$ from this we get $$\tan \psi=\frac{mg}{T_0}\Rightarrow \frac{dm}{d\psi}=\frac{T_0}{g}\sec^2\psi$$

Putting these results together, we get $$\frac{ds}{d\psi}=\frac{ds}{dm}\frac{dm}{d\psi}=a\sec^3\psi,$$ where $a=\frac{T_0^2}{\rho k}$ is a comstant.

We now have an expression for the radius of curvature of the curve which is a parabola with its vertex vertical, as can be verified quite easily as follows.

Taking initial conditions $x=0, y=0, \psi=0$, $$x=\int\frac{ds}{d\psi}\cos\psi d\psi=\int a\sec^2\psi d\psi=a\tan\psi$$ And $$y=\int\frac{ds}{d\psi}\sin\psi d\psi=\int a\sec^2\psi \tan\psi d\psi=\frac 12a\tan^2\psi$$

Hence$$y=\frac{x^2}{2a}$$ QED