**(a) In equilibrium, kx**_{0} = mg....(i)

**When further depressed by an amount x, net restoring force (upwards) is,**

**(b) In this case if the mass m moves down a distance x from its equilibrium position, then pulley will move down by x/2. So, the extra force in spring will be kx/2. Now, as the pulley is massless, this force kx/2 is equal to extra 2T or T = kx/4. This is also the restoring force of the mass. Hence,**

**(c) In this situation if the mass m moves down a distance x from its equilibrium position, the pulley will also move by x and so the spring will stretch by 2x. Therefore the spring force will be 2kx. The restoring force on the block will be 4kx. Hence,**