A stone of mass m is tied to an elastic string of negligible mass...

A stone of mass m is tied to an elastic string of negligible 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. Find the distance y from the top when the mass comes to rest for an instant, for the first time

Options

A $y=\frac{\left( kL-mg \right)\pm \sqrt{mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
B $y=\frac{\left( kL+mg \right)\pm \sqrt{mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
C $y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
D $y=\frac{\left( kL-mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$

Correct answer

C. $y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$

Step-by-step solution

Correct Option is : (C)

y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}

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