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I wanted to ask why a pulse that reaches to a free end will reflect off and return with the same direction of displacement that it had before reflection. I am looking for explanation with forces.

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  • $\begingroup$ Related question for electromagnetic oscillations. Or are you talking about a wave on a string? Please clarify. $\endgroup$
    – rob
    May 23, 2014 at 16:52
  • $\begingroup$ a wave on a string $\endgroup$
    – Michael
    May 23, 2014 at 16:53
  • $\begingroup$ For a wave moving on a string, pulses generally don't reflect from a free end. Think of the way that a "pulse" moves down a bullwhip and dissipates into the atmosphere at the end. $\endgroup$
    – rob
    May 23, 2014 at 17:00
  • $\begingroup$ en.wikipedia.org/wiki/Pulse_(physics) - Pulse Reflection , Free End $\endgroup$
    – Michael
    May 23, 2014 at 17:24
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    $\begingroup$ @rob in a strict sense you may be right, but free end in such contexts typically means free in one dimension only, as in the latter half of your answer below. $\endgroup$
    – BMS
    May 23, 2014 at 18:12

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For a wave moving on a string, you don't get a reflection off of a completely free end. Think of the way that a "pulse" moves down a bullwhip and dissipates into the atmosphere at the end.

If the position of the end of the string is fixed in three dimensions, there's a symmetry argument that says you have to get a reflected pulse with the opposite displacement from the incident pulse: at the fixed point, the incident pulse and the reflected pulse must add up to give zero displacement. This gives you a reflection with the same shape as the incident pulse but the opposite sign for the displacement and moving in the opposite direction. The forces involved here are the tension in the string and the restoring force of whatever fixture holds the end of the string in place. In the language on differential equations this is usually called a "Dirichlet boundary condition."

The example you linked has what I might call a "nearly free end": the end is free to be displaced transverse to the string, but its position along the string is fixed. If you wanted to build one of these, you might tie your string to a ring, and put the ring around a fireman's pole. Now the end of the string can't move towards you, like the end of a whip does, but it's free to move up and down. (If you don't have a two-story fire station handy, you could use a shower curtain rod. You might have to use a heavy-ish rope so that the mass of the ring can be neglected.)

If there's not any friction between the ring and the pole, and there's healthy tension in the string, you'll notice something funny happen: the string will always make a right angle to the pole! If the string does make an angle with the pole, the tension in the string will move the ring up or down until the string is normal to the pole again. So here we have the derivative of the displacement being fixed at the end of the string. This is called a Neumann boundary condition.

You can use the same sort of symmetry argument as with the Dirichlet boundary to find the shape of the reflected pulse. You know that the slope of the incoming pulse will be exactly canceled out by the slope of the reflection. The solution is that the incident and reflected pulses have the same shape and the same sign, but are moving in opposite directions. Here the only force involved is the tension in the string, and the frictionless condition that the end is "nearly fixed."


This model of a rope and a ring and a pole is perhaps not completely realistic. However Neumann boundary conditions have applications to lots of other oscillators. In particular, in electromagnetism, we have the boundary condition that the electric field is always perpendicular to the surface of a conductor — or equivalently, that the component of the electric field parallel to the surface of a conductor is always zero — which has the same sort of counterfactual justification: if there were an electric field parallel to the surface of a conductor, the free charges on the conductor would feel pushed around and move, until the field was cancelled out.

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  • $\begingroup$ Could you elaborate and explain why the slope of the incoming pulse will be exactly canceled out by the slope of the reflection? $\endgroup$
    – Michael
    May 24, 2014 at 7:11
  • $\begingroup$ @Michael Any better? $\endgroup$
    – rob
    May 24, 2014 at 14:55

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