In physics textbook the potential energy of a segment is often given by
$$\Delta U=F(dl-dx). $$
I know this is the work done to stretch the string. However, shouldn't we consider the work done that move this segment up and down under the action of wave?
Explicitly,
$$\Delta U=\int F_y ~dy $$
where $F_y$ is net force exerted on the segment.
You can't get that work back. i.e. it is not potential energy, it is loss.
Potential energy is energy stored in a string, just as gravitational potential energy is energy stored due to height. When the string stretches elastically the energy goes in, when it contracts the energy comes out again. When you travel downward gravitational potential energy is released to increase kinetic energy.
Within each oscilation as the string element moves laterally, as it has mass this takes work. It has to speed up, taking work, slow down to zero, reverse speed. When the oscilation is maximum amplitude the string is stationary (imagine a standing wave). There is no kinetic energy, all the energy is in the elasticity of the string. When the string is at minimum amplitude all the energy is kinetic, and the string is contracted. The energy is constantly oscilating between elastic potential energy and kinetic energy. Kinetic movement of the mass will encounter the usual losses, friction, air-ristance. It's a source of loss.
