Sarthaks Test
0 votes
in Physics by (8.6k points)

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?

1 Answer

0 votes
by (13.6k points)
selected by
Best answer

(a) 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 a SHM. Let the stone come to rest instantaneously at y.

The loss in P.E. of the stone is the P.E. stored in the stretched string.

(b) The maximum velocity is attained when the body passes, through the “equilibrium, position” i.e. when the instantaneous acceleration is zero. That is mg - kx = 0 where x is the extension from L:

(c) Consider the particle at an instantaneous position y. Then

Make a transformation of variables: z = k/m(y - L) - g

Thus the stone performs SHM with angular frequency ω about the point

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.