# Mechanics ​Q6. One end of a light spring of natural length d and spring constant k (= mg/d) is fixed on a rigid support and the other end is fixed to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the support. Initially, the spring makes an angle of 37° with the horizontal as shown in the figure.The system is released from rest. The speed of the ring at the same angle subtended downward will be          (A) $\sqrt{2\mathrm{gd}}$                                                     (B) $\sqrt{3\mathrm{gd}}$         (C) $\sqrt{4\mathrm{gd}}$                                                      (D)

Dear Student

Cos37= h/l
l= h/cos37
l=1.25h
extension = l-h
=1.25h- h= 0.25h

equalling KE and PE
1/2mv2= 1/2k(0.25h)2
v2= k/m x h2/16
v= h/4 √k/m

Regards

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