0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Comment to the post (v2): Is $x$ and $y$ supposed to be perpendicular coordinates? $\endgroup$ – Qmechanic Dec 23 '16 at 2:54
  • $\begingroup$ Yes in the usual sense $\endgroup$ – Math The Novice Dec 23 '16 at 3:37
  • $\begingroup$ Combined transversal and longitudinal displacements of the non-relativistic string are worked out in my Phys.SE answer here. $\endgroup$ – Qmechanic Dec 25 '16 at 21:57
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This does not appear to attempt to answer the question. $\endgroup$ – Kyle Oman Dec 22 '16 at 19:39
  • $\begingroup$ I don't quite understand what you mean here. Can you elaborate why it is loss not potential? $\endgroup$ – Math The Novice Dec 23 '16 at 1:43
  • $\begingroup$ Potential energy is energy stored in a string, just a sgravitational 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. $\endgroup$ – JMLCarter Dec 23 '16 at 1:54
  • $\begingroup$ When we calculate gravitational potential we use $F={GMm \over r^2}$ Can we use same reasoning that $F_y=T{d^2y \over dx^2}dx ? So the total potential would be this plus the potential of elongation $\endgroup$ – Math The Novice Dec 23 '16 at 3:44
  • $\begingroup$ It's just simpler to know that when the waves are at maximum amplitude there is no kinetic energy in the string it is stationary as it changes direction. All energy is in the stretched string. (The amount of energy doesn;t alter when some of it becomes kinetic energy.) That way you don't need to add a term for the physical movement of the wave. $\endgroup$ – JMLCarter Dec 23 '16 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.