Question

A stone of mass m is tied to an elastic string of negligble mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.
(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop?
(c) What shall be the nature of the motion after the stone has reached its lowest point?

Answer 1

$\left(a\right)$Till the stone drops through a length $L$ it will be in free fall. After that

`the elasticity of the string will force it to S.H.M.Let the stone come to rest

instantaneously at $y$

$\begin{array}{c}\therefore mgy=\frac{1}{2}R{\left(y-L\right)}^2\\ or,mgy=\frac{1}{2}R{y}^2-RyL+\frac{1}{2}R{L}^2\\ or,y=\frac{\left(RL+mg\right)\pm \sqrt{{\left(RL+mg\right)}^2-R^2{L}^2}}{R}\\ or,y=\frac{\left(RL+mg\right)\pm \sqrt{2mgRL+m^2{g}^2}}{R}\end{array}$

$\left(b\right)$For maximum velocity instantaneous acceleration is $0$

$mg-Rx=0$ where x is extension from $L$

Let the velocity be $V$

$\begin{array}{c}\therefore \frac{1}{2}m{v}^2+\frac{1}{2}R{x}^2=mg\left(L+x\right)\\ Now,mg=Rx\\ x=\frac{mg}{R}\\ \therefore \frac{1}{2}m{v}^2=mg\left(L+\frac{mg}{R}\right)-\frac{1}{2}R\frac{m^2{g}^2}{R^2}\\ \therefore V^2=2gL+\frac{mg}{R}\\ V={\left(2gL+\frac{m{g}^2}{R}\right)}^{1/2}\end{array}$

$\left(c\right)$Consider the particle at an instantaneous position $y$

$\begin{array}{c}\frac{m{d}^{2}y}{d{t}^2}=mg-k\left(y-l\right)\\ or,\frac{d^{2}y}{d{t}^2}+\frac{k}{m}\left(y-L\right)-g=0\\ Letz=\frac{k}{m}\left(y-L\right)-g\\ \therefore \frac{d^{2}z}{d{t}^2}+\frac{R}{m}z=0\\ \therefore z=A\cos \left(wt+\varphi \right)wherew=\sqrt{\frac{R}{m}}\\ or,y=\left(L+\frac{m}{R}g\right)+A^{{}^{\prime }}\cos \left(wt+\varphi \right)\end{array}$

Thus, the stone perform S.H.M with angular frequency $w$ about the point

$y_0=L+\frac{m}{R}g$