One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is fixed to a smooth ring of mass m which can slide without friction in 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 the diagram. When the system is released from rest, find the speed of the ring when the spring becomes horizonta

Step-by-step solution

If l ⁡ is the stretched length of the spring d l ⁡ = cos 37 o = 4 5 , i.e., l ⁡ = 5 4 d So, the stretch y = l ⁡ - d = 5 4 d - d = d 4 and AB = h = l ⁡ sin 37 o = 5 4 d × 3 5 = 3 4 d Now taking point B as a reference level and applying the law of conservation of Energy between A and B , E A = E B mgh + 1 2 ky 2 + 0 = 0 + 0 + 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 3g 2d + k 16m