Question

A ring of mass m is attached to a horizontal spring of spring constant K and natural lenght L0,other end of spring is fixed and ring can slide on a smooth horizontal rod.Now the ring is shifted to position B and released,speed of ring when spring attains it's natural lenght is-

Answer 1

your question --> 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 as shown in figure . Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical ?

solution : h = Lo = natural length

from ∆ABC,

cos37° = BC/AC = h/(h + x) , here x is extension of spring when it stays at point A.

or, cos37° = 4/5 = h/(h + x)

or, 4h + 4x = 5h => x = h/4 ......(1)

now applying work energy theorem,

spring potential energy = kinetic energy

or, 1/2 kx² = 1/2 mv²

or, v² = kx²/m

or, v = x√(k/m) = h/4√(k/m) [ from equation (1) ]

hence, speed of ring when spring attains its natural length or becomes vertical is $\frac{h}{4}\sqrt{\frac{k}{m}}$