Question

Consider a rope of total mass $M$ and length $L$ suspended at rest from a fixed mount. The rope has a linear mass density that varies with height as $\lambda \left(z\right)={\lambda }_0\sin (\pi z\setminus L)$, where ${\lambda }_0$ is a constant. Constant gravitational acceleration $g$ acts downward.

The tension along the rope as a function of distance $z$ below the mount is

• A. $T=\frac{Mg}{2}[1+\cos \frac{\pi z}{L}]$
• B. $T=Mg[1+\cos \frac{\pi z}{L}]$
• C. $T=\frac{Mg}{2}[1+\cos \frac{\pi }{L-z}]$
• D. $T=\frac{Mg}{2}[1-\cos \frac{\pi z}{L}]$

Answer 1

The correct option is A $T=\frac{Mg}{2}[1+\cos \frac{\pi z}{L}]$


$T=mg$ ....(i)
We have been given that
$\lambda ={\lambda }_0\sin \left(\frac{\pi z}{L}\right)$
$\Rightarrow \frac{dm}{dz}={\lambda }_0\sin \left(\frac{\pi z}{L}\right)$
$\Rightarrow {\int }_0^{m}dm={\int }_0^L{\lambda }_0\sin \left(\frac{\pi z}{L}\right)dz$
$\Rightarrow M=\frac{{\lambda }_{0}L}{\pi }[-\cos \left(\frac{\pi z}{L}\right){]}_0^L$ ...(ii)
$\Rightarrow M=\frac{{\lambda }_{0}L}{\pi }[1+1]$
$\Rightarrow {\lambda }_0=\frac{\pi M}{2L}$
In eq(ii) we just need to alter the mass M to m and the limts from $0\to L$ to $0\to (L-z)$
we get
$m=\frac{{\lambda }_{0}L}{\pi }[-\cos \left(\frac{\pi z}{L}\right){]}_0^{L-z}$
$\Rightarrow m=\frac{\pi M}{2L}\frac{L}{\pi }[1-\cos (\frac{\pi L}{L}-\frac{\pi z}{L})]$
$m=\frac{M}{2}[1+\cos \frac{\pi z}{L}]$
Putting this in eq(i)
$T=\frac{Mg}{2}[1+\cos \frac{\pi z}{L}]$