One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of `37^(@)` with the horizontal as shown in fig. When the system is released from rest, find the speed of the ring when the spring becomes horizontal.
[ sin `37^(@)` = 3/5]

Explanation

If l is the stretched length of the spring, then from figure
`(d)/(l) = cos 37^(@) = (4)/(5) , " " i.e., " " l = (5)/(4) d `
So, the stetch y = l - d = `(5)/(4) d -d = (d)/(4)`
and h = l sin `37^(@) = (5)/(4) d xx (3)/(5) = (3)/(4) d `
Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B,
`E_(A) = E_(B)`
or `" mgh"+ (1)/(2) ky^(2) = (1)/(2) mv^(2)`
[ as for, B, h = 0 and y = 0 ]
or `" " (3)/(4) "mgd" + (1)/(2) k ((d)/(4))^(2) = (1)/(2) mv^(2)`
[ as for A, h = `(3)/(4)` d and y = `(1)/(4) ` d ]
or `" " v = d sqrt((3g)/(2d) + (k)/(16m))`