Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.
The spring will be stretched to balance the tension in the string.
Let the elongation be x0
⇒ Kx0 = T0
⇒ 2 T0 = mg
⇒ 2 Kx0 = mg ....(1)
Now if I pull the pulley by x downwards then there is an extra AB = C D = X part in the string.
⇒ a total of 2x elongation. Since the string is non stretchable so obviously spring is further elongated by 2x on top of the x0 at equilibrium. So Now the spring force is K (2 x + x0) which will be equal to the tension in the strings.
K(2x + x0) = T ---------------(ii)
⇒ 2T - mg = ma
2k(2x + x0) - 2Kx0 = ma
from equation (i) & (ii)
a = 4kxm