An elastic cord of elastic constant k and length L is hung from point A having a massless block at the other end. A smooth ring of mass M falls from point A. What is the maximum elongation of cord?
To find the maximum elongation of the cord, we need to consider the equilibrium of forces acting on the system.
1. Let's assume the maximum elongation of the cord is x.
2. The weight of the smooth ring is acting downwards, which can be calculated as W = Mg, where M is the mass of the ring and g is the acceleration due to gravity.
3. The tension in the cord will increase as the ring falls, trying to restore equilibrium. At equilibrium, the tension in the cord would be equal to the weight of the ring, so we can write T = Mg.
4. The elastic force in the cord is given by Hooke's Law, F = -kx, where k is the elastic constant of the cord, and x is the elongation of the cord.
5. Since the cord is massless, at equilibrium, the net force acting on the system is zero, so we can write the equation as:
0 = T - W - F
Substituting the values of T, W, and F:
0 = Mg - Mg - (-kx)
0 = -kx
6. Solving the equation, we get:
kx = 0
Since the elastic constant k cannot be zero, the only possibility is:
x = 0
Therefore, the maximum elongation of the cord is zero. This means that the cord does not stretch or elongate when the ring is falling from point A.