Question

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]

Answer 1

Correct Answer - A::B::C::D
If l is the stretched length of the spring, then figure
`d/l=cos 37^(@) =4/5, ` i.e. `l=5/4d`
So, the stretch `x=l-d =5/4d-d =d/4`
and `h=l` sin `37^(@) =5/4d xx 3/5 =3/4d`
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 +1/2kx^(2) =1/2mv^(@)` , [At `B, h=0 and x =0`]
or `3/4mgd + 1/2k(d/4)^(2) =1/2mv^(2)` [as for `A, h=3/4d` and `y=1/4d`]
or `v=d sqrt((3g)/(2d)+k/16m))`