one end of a spring of naturla length ha and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is alowed to slide on a horizontal rod fixed at a height h figure. Initialy, the spring makes an angle of `37^0` with the vertical when teh system is released from rest. find teh speed of the ring when teh spring becomes vertical.

Answer

D

Explanation

`theta=37^0,l=h` N/Atural length
Let the velocity be v

`cos37^0=(BC)/(AC)=0.8=4/5`
`A_c=(h+x)=(5h)/4`
Applying work energy principle
`1/2kx^2=1/2 mv^2`
`rarr=x sqrt((k/m))=h/4 sqrt((k/m))`.