A small block of mass m is placed on a bigger block of mass M that is attached to a vertical spring of spring constant k . The system oscillates vertically. Find the normal force on the smaller block when it is displaced by a distance x above its equilibrium position. Also, identify when this force is smallest in magnitude.

Step-by-step solution

Let the upward direction be taken as positive. When the system is displaced by a distance x above its equilibrium position, the restoring force acting on the combined mass (M+m) is given by: F = -kx The acceleration of the system is: a = F M+m = - kx M+m Considering the free body diagram of the smaller block of mass m , the forces acting on it are the normal force N upwards and its weight mg downwards. Applying Newton's second law for the smaller block: N - mg = ma Substituting the expression for acceleration a : N