Two children in a school corridor tie one end of a skipping rope to a door handle. One of them flicks the free end once. A single hump runs along the rope to the handle, and a hump runs back. It looks like a ball bouncing off a wall. But a hump is not a thing. It is a shape, made by pieces of rope that only move up and down.
A reflection is not a bounce
When a wave travels along a string, we often picture it bouncing off a boundary. But what is actually happening? The truth is more elegant. A reflection is not a bounce. It is the creation of a new wave. It is a new action that the system takes, so that the fundamental laws of physics are never broken at the boundary.
The state of a wave
At any point of a uniform string, the state of the wave is completely described by two linked properties.
- The displacement \(y\): how far the string is from its rest position.
- The transverse momentum: how fast the string is moving up or down. For a piece of mass \(\mu\,dx\), it is \(\mu\,dx\,(\partial y/\partial t)\), where \(\mu\) is the linear mass density. It is directly related to the slope \(\partial y/\partial x\) of the string.
The chapter Introduction to Waves gave that relation for a wave \(y = f(x - vt)\) moving at speed \(v\): \(v_{\text{particle}} = \partial y/\partial t = -v\,\partial y/\partial x\). So in a single travelling wave, the velocity of each piece is fixed by the slope there.
These two properties are not just descriptions. They are the initial conditions of the wave. Suppose we could freeze time and give the displacement and the transverse momentum at every point of the string. Then the whole future of the wave would be fixed. Only one motion could follow.
What a boundary does
In a uniform medium, displacement and slope are linked naturally. A boundary is different. It is a special point that imposes an outside rule that cannot be broken. The rule forces a fixed value on one of the two properties. Such a rule is a boundary condition.
- A rigid boundary forces the displacement to be zero, \(y = 0\), whatever the slope does.
- A free boundary forces the slope to be zero, \(\partial y/\partial x = 0\), whatever the displacement does.
A boundary condition is a rule that a boundary imposes on the string at all times. At a rigid boundary the displacement is always zero. At a free boundary the slope is always zero.
In Fig 3.1 the rigid end is a string tied to a wall. The wall cannot move, so the end of the string cannot move. The free end is a ring on a smooth vertical rod. The ring may slide to any height, but the string always meets the rod at right angles. The dashed line through the ring shows that the string is level there.
When an incident wave arrives, it tries to make the boundary behave like any other point of the string. That would break the boundary's special rule, and the system cannot allow it. Its only solution is to generate a new disturbance at the boundary.
A string is tied firmly to a steel pillar. Which quantity is forced to be zero at the tied end at every instant?
The end of a string is tied to a light ring that slides without friction on a smooth vertical rod. What does this boundary force?
The new disturbance
A boundary imposes a rule that the incoming wave, \(y_{\text{incident}}\), tries to break. The system's only way out is to generate an additional disturbance at the boundary. Call it \(y_{\text{new}}\). It is there to make sure the rule is obeyed. The actual displacement of the string, \(y_{\text{string}}\), is therefore the superposition of the two:
\[ y_{\text{string}} = y_{\text{incident}} + y_{\text{new}} \]
Now use a fundamental principle: the physical string must obey the wave equation at all points and at all times. The wave equation comes from Newton's laws applied to each piece of the string, and those laws have no exceptions. So \(y_{\text{string}}\) must be a solution of the wave equation, and the steps below show what follows.
\[ y_{\text{string}} = y_{\text{incident}} + y_{\text{new}} \]
Why. the principle of superposition: the string moves with the sum of the incident wave and the new disturbance.
\[ \dfrac{\partial^2 y_{\text{string}}}{\partial x^2} = \dfrac{1}{v^2}\dfrac{\partial^2 y_{\text{string}}}{\partial t^2} \]
Why. the physical string obeys the wave equation at every point and every instant, because it obeys Newton's laws.
\[ \dfrac{\partial^2 (y_{\text{incident}} + y_{\text{new}})}{\partial x^2} = \dfrac{1}{v^2}\dfrac{\partial^2 (y_{\text{incident}} + y_{\text{new}})}{\partial t^2} \]
Why. substitute the superposition of step 1 into the wave equation of step 2.
\[ \left(\dfrac{\partial^2 y_{\text{incident}}}{\partial x^2} - \dfrac{1}{v^2}\dfrac{\partial^2 y_{\text{incident}}}{\partial t^2}\right) + \left(\dfrac{\partial^2 y_{\text{new}}}{\partial x^2} - \dfrac{1}{v^2}\dfrac{\partial^2 y_{\text{new}}}{\partial t^2}\right) = 0 \]
Why. differentiation is linear: the derivative of a sum is the sum of the derivatives. So the terms can be separated and grouped into two brackets.
\[ \dfrac{\partial^2 y_{\text{incident}}}{\partial x^2} - \dfrac{1}{v^2}\dfrac{\partial^2 y_{\text{incident}}}{\partial t^2} = 0 \]
Why. the incident disturbance is a wave, so by definition it satisfies the wave equation. The first bracket is zero.
\[ \dfrac{\partial^2 y_{\text{new}}}{\partial x^2} - \dfrac{1}{v^2}\dfrac{\partial^2 y_{\text{new}}}{\partial t^2} = 0 \]
Read it. step 4 says the two brackets add to zero, and step 5 says the first is zero. So the second must be zero too.
This is a powerful conclusion. We have proved that the new disturbance must itself be a wave. It is generated at the boundary, and it travels back into the original medium. So we can now give it its familiar name: the reflected wave.
Near a boundary, the string is the sum of the incident wave and a new disturbance created by the boundary. Because the string and the incident wave both obey the wave equation, the new disturbance obeys it too. It is a wave travelling away from the boundary: the reflected wave.
Watch Fig 3.2 as the incident pulse reaches the wall. The new disturbance grows out of the wall just as the incident pulse runs into it. The readout gives the two at the wall: they are always equal and opposite, so their sum there is zero. Once the incident pulse has gone, only the new disturbance is left on the string. It moves away from the wall like any other pulse. At this rigid end it is upside down; the next lesson proves why.
Think of the boundary as a guard with one order. At a wall the order is "this point stays at zero". An arriving wave tries to move the point. So the guard answers with a wave of its own, exactly enough to cancel the push at that point. The guard's answer cannot stay at the wall. It is a wave, so it travels.
Put the steps of the proof that the new disturbance is a wave in order, first step first.
Why must the string obey the wave equation even at the instant a pulse reaches a wall?
The wave equation is Newton's second law applied to each small piece of the string, with the tension supplying the force. Newton's laws hold at every instant, including the moment of reflection. So every piece of the string still accelerates as the wave equation says. The wall only adds a rule at one point, the end.
A student says: "The reflected pulse is the incident pulse that bounced off the wall." What is wrong with this picture?
In 1747 the French mathematician d'Alembert wrote down the wave equation for a vibrating string. He then found every one of its solutions. Each is the sum of two shapes, \(f(x - vt) + g(x + vt)\), one sliding to the right and one to the left. So any disturbance that obeys the wave equation is made of travelling waves. The new disturbance at a boundary is one of them, the wave \(g(x + vt)\) moving back.
Now return to the rope on the door handle. The hump that comes back is not the first hump returning. The handle holds its end still, and to keep that rule it creates a new disturbance. The rope obeys the wave equation, so that disturbance is a wave, and it runs back along the rope.
- Picturing reflection as the incident pulse bouncing like a ball.The boundary creates a new wave to keep its rule. The incident pulse runs into the boundary and is replaced.
- Thinking a boundary fixes both the displacement and the slope.A rigid end fixes only \(y = 0\). A free end fixes only \(\partial y/\partial x = 0\).
- Believing the wave equation fails at the moment of reflection.The string obeys Newton's laws at every instant. The boundary adds one rule at one point.