Question

A rope of uniform mass $m$ and length $L$ is hanging from a support. A block of mass $M$ is attached to the rope at the lower end. If a transverse pulse is produced at the lower end, find the time taken by the pulse to reach the upper end.

• A. $2\sqrt{\frac{L}{g}}[\sqrt{(1+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$
• B. $\frac{3}{2}\sqrt{\frac{L}{g}}[\sqrt{(2+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$
• C. $\sqrt{\frac{L}{g}}[\sqrt{(1+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$
• D. $\frac{3}{2}\sqrt{\frac{L}{g}}[\sqrt{(1+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$

Answer 1

The correct option is A $2\sqrt{\frac{L}{g}}[\sqrt{(1+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$


Linear mass density of rope $\left(\mu \right)=\frac{m}{L}$
Tension at point A, $T_A=\mu xg+Mg$
Velocity of transverse pulse at point A
$v=\sqrt{\frac{T_A}{\mu }}=\sqrt{\frac{(\mu x+M)g}{\mu }}$
$\Rightarrow \frac{dx}{dt}=\sqrt{(x+\frac{M}{\mu })}\sqrt{g}[\because v=\frac{dx}{dt}]$
To find time taken for wave pulse to travel to the upper end, integrate both sides setting the limit of $x$ from $0$ to $L$
$\Rightarrow {\int }_0^L\frac{dx}{\sqrt{(x+\frac{M}{\mu })}}=\sqrt{g}{\int }_{t=0}^{t}dt$
$\Rightarrow 2{\left[\sqrt{(x+\frac{M}{\mu })}\right]}_0^L=\sqrt{g}[t{]}_0^t$
$\Rightarrow 2[\sqrt{(L+\frac{M}{\mu })}-\sqrt{\frac{M}{\mu }}]=\sqrt{g}t$
$\Rightarrow t=\frac{2[\sqrt{(L+\frac{M}{\mu })}-\sqrt{\frac{M}{\mu }}]}{\sqrt{g}}=\frac{2\sqrt{L+\frac{ML}{m}}-2\sqrt{\frac{ML}{m}}}{\sqrt{g}}[\because \mu =\frac{M}{L}]$
$\Rightarrow t=2\sqrt{\frac{L}{g}}[\sqrt{(1+\frac{M}{m})}-\sqrt{\frac{M}{m}}]$