Question

A uniform rope of mass $0.1 \,kg$ and length $2.45 \,m$ hangs from a ceiling. The time taken by a transverse wave to travel the full length of the rope is $\left(g=9.8 \,m / s ^{2}\right)$

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Answer 1

Answer: (A)

As the rope has mass and it is suspended vertically, tension in it will be different at different points. For a point at a distance $x$ from the free end, tension will be due to the weight of the rope below it. If $M$ is the mass of rope of length $L$, the mass of length $x$ of the rope will be $(M / L) x$.
$\therefore \,\,\,\,T=\left[\frac{M}{L} x\right] g$
and $v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{M g x}{L(M / L)}}=\sqrt{g x}\,\,\,\,\,\,\,\,\,...(i)$
The tension and the velocity of the wave is different at different points. If at point $x$ the wave travels a distance $d x$ in time $d t$,
$v=\frac{d x}{d t} \text { or } \sqrt{g x}=\frac{d x}{d t}$ (Using(i))
or $\int d t=\int \frac{d x}{\sqrt{g x}}$ or $t=\frac{1}{\sqrt{g}} \int\displaystyle_{0}^{L} x^{-1 / 2} d x$
or $t=2 \sqrt{\left(\frac{L}{g}\right)}$
As $L=2.45\, m$
$ \therefore t=2 \sqrt{\frac{2.45}{9.8}}=1\, s$