question_answer 50)

                  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 die 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:

                  (a) P.E. of stone at the height \[y=mgy\]                 When stone falls through height \[y,\] then                 Elastic potential energy stored in the spring                 \[=\frac{1}{2}k{{(y-L)}^{2}}=mgy\]                 Using the value of conservation of energy                 \[\frac{1}{2}k{{(y-L)}^{2}}=mgy\]                 or \[k{{(y-L)}^{2}}=mgy\]                 or \[k{{(y-L)}^{2}}=2mgy\]                 or \[k({{y}^{2}}+{{L}^{2}}-2yL)=2mgy\]                 or \[k{{y}^{2}}-2y(kL+mg)+k{{L}^{2}}=0\]                 \[\therefore \] \[y=\frac{2(kL+mg)\pm \sqrt{4{{(kL+mg)}^{2}}-4{{k}^{2}}{{L}^{2}}}}{2k}\]                 \[=\frac{(kL+mg)\pm \sqrt{{{(kL+mg)}^{2}}-{{k}^{2}}{{L}^{2}}}}{k}\]                 Since \[y\] is positive, therefore,                 \[y=\frac{(kL+mg)+\sqrt{{{(kL+mg)}^{2}}-{{k}^{2}}{{L}^{2}}}}{k}\]                 (b) Let \[x\] be the extension in the spring when stone drops, then at the equilibrium position.                 \[mg(L+x)=\frac{1}{2}m{{\upsilon }^{2}}+k{{x}^{2}}\]                 Also \[mg=kx\] or \[x=\frac{mg}{k}\]                 \[\therefore \] \[mg\left( L+\frac{mg}{k} \right)=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\frac{{{(mg)}^{2}}}{k}\]                 or \[\frac{1}{2}m{{\upsilon }^{2}}=mgL+\frac{{{(mg)}^{2}}}{k}-\frac{{{(mg)}^{2}}}{2k}\]                 or \[\frac{1}{2}m{{\upsilon }^{2}}=mgL+\frac{{{(mg)}^{2}}}{2k}\]                 or \[\upsilon =\sqrt{2gL+\frac{m{{g}^{2}}}{R}}\]