Question

A uniform string of length L and total mass M is suspended vertically and a transverse pulse is given at the top end of it. At the same moment a body is released from rest and falls freely from the top of the string. If the body passes the pulse at a distance $L/x$ from the bottom. Find $x$.

Answer 1

Let the pulse given at the bottom of the string and the body meet at the line $AP$ after time $t$.

Thus the pulse travels a distance $y$ and the body falls by distance $L-y$.

To find velocity of pulse when it is at distance $y$ from bottom :

Let mass per uni length of the string be $\mu$

Tension in the string below line OA $T=\mu yg$

$\therefore$ Velocity of the pulse $v=\sqrt{\frac{T}{\mu }}=\sqrt{yg}$

$\therefore$ $v=\frac{dy}{dt}=\sqrt{yg}$

Integrating, we get ${\int }_0^y{y}^{-1/2}dy=\sqrt{g}{\int }_0^{t}dt$

$\frac{y^{1/2}}{\frac{1}{2}}{|}_0^y=\sqrt{g}t{|}_0^t$ $⟹\sqrt{y}=\frac{1}{2}\sqrt{g}t$ OR $y=\frac{1}{4}g{t}^2$ ........(1)

For freely falling body : $S=ut+\frac{1}{2}g{t}^2$

$\therefore$ $L-y=0+\frac{1}{2}g{t}^2$ $⟹L-y=\frac{1}{2}g{t}^2$ ........(2)

From (1) and (2), $L-\frac{1}{4}g{t}^2=\frac{1}{4}g{t}^2$

$⟹$ $L=\frac{3}{4}g{t}^2$ ..........(3)

Now as $y=\frac{L}{x}$ $⟹x=\frac{L}{y}$

$\therefore$ $x=\frac{\frac{3}{4}g{t}^2}{\frac{1}{4}g{t}^2}=3$