The potential energy of a spring pendulum, with a mass m connected to it, is given by `V=1/2kx^(2)` (where x = displacement from the position of equilibrium and k = a constant). How does the applied force on the mass vary with displacement?

Explanation

Suppose, the force applied on the mass m = F.
If V is the potential energy of the spring pendulum, then
`F=-(dV)/(dx)=-d/(dx)(1/2kx^(2))=-1/2k*2x=-kx`
So, the applied force F is proportional to x and acts in a direction opposite to that of x.