Question

One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (Figure) Initially, the spring makes an angle of 37° with the vertical when the system is released from the rest. Find the speed of the ring when the spring becomes vertical.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question :- 
θ = 37°
l =h = natural length.
Initial position length of the spring = hsec37°
Elongation in the spring, x  = hsecθ -h  .
x = 125h -h
x =0.25h
. Now potential energy stored in the spring = 1/2kx² 
=1/2 k(0.25h)² .
Hence when the spring comes to it's vertical position stored potential energy & initial kinetic energy = 0.
Now total Potential energy is converted to Kinetic energy.

1/2 mv² = 1/2k(0.25h)²
$v = x \sqrt{k/m}$ 
$v = h/4 \sqrt{k/m}$ 
Hence the spped of the ring when spring become vertical is $v = h/4 \sqrt{k/m}$ .


Hope it Helps.