A teacher lays a long spring, a slinky, along a smooth floor. First she clamps its far end to a table leg and flicks the near end sideways. The pulse runs to the clamp and comes back on the other side. Then she ties the far end to a long, light thread instead. Now the pulse comes back on the same side, and the free end whips out a long way.
At a rigid end the rule is that the total displacement of the string at the end must be zero, at every instant:
\[ y_{\text{string}}(\text{end}, t) = y_{\text{incident}} + y_{\text{reflected}} = 0 \]
To enforce this rule, the system generates a reflected wave that is the exact opposite of the incident wave at the boundary:
\[ y_{\text{reflected}}(\text{end}, t) = -\,y_{\text{incident}}(\text{end}, t) \]
This means that the reflected wave must be an inverted copy of the incident wave. At the boundary it interferes destructively with the incident wave, which keeps the end stationary. Then it travels back along the string. This is what we observe as a reflection with a phase change of \(\pi\) radians, or \(180°\).
Press play in Fig 4.1. The pulse, \(0.8\,\text{cm}\) high, starts at \(x = 3\,\text{m}\) and runs at \(2\,\text{m/s}\) to the wall at \(10\,\text{m}\). It reaches the wall at \(3.5\,\text{s}\). The readout of the tied end stays at \(0.00\,\text{cm}\) the whole time. By \(7\,\text{s}\) an inverted pulse of the same size is back at \(x = 3\,\text{m}\).
A wave that obeys the rule
We can write the reflected wave down. Let the incident wave be \(y_{\text{incident}} = f(x - vt)\), and let the fixed end be at \(x = L\). The derivation checks a candidate.
\[ y_{\text{reflected}}(x, t) = -\,f(2L - x - vt) \]
Why. take the incident shape, mirror it about the end, and turn it upside down.
\[ 2L - x - vt = 2L - (x + vt) \]
Why. the argument depends on \(x\) and \(t\) only through \(x + vt\), so this is a wave of the form \(g(x + vt)\), travelling towards \(-x\) at speed \(v\).
\[ y_{\text{reflected}}(L, t) = -\,f(L - vt) = -\,y_{\text{incident}}(L, t) \]
Why. put \(x = L\) in step 1. The result is the fixed-end rule, obeyed at every instant.
\[ y_{\text{string}} = f(x - vt) - f(2L - x - vt) \]
Read it. superpose the two waves. This is the string on \(x \le L\), and it is how Fig 4.1 is drawn.
For a harmonic wave the inversion is a shift of phase. An inverted sine is the same sine half a cycle on, since \(-A\sin\theta = A\sin(\theta + \pi)\). That is why a fixed end is said to add a phase change of \(\pi\).
The candidate of step 1 is called an image. Imagine the string continued beyond the wall. An inverted pulse approaches from the far side, as far behind the wall as the real pulse is in front of it. The two meet at the wall and cancel there at every instant. On the real string, the image pulse is the reflected pulse. This is the method of images, and it is used in electrostatics too.
Case 2: the free boundary
At a free end the string ends in a light ring on a smooth vertical rod. The ring is taken as massless, so the net vertical force on it must be zero; otherwise it would have an infinite acceleration. The rod is smooth and can push only sideways. So the vertical pull of the string on the ring must be zero. That pull is proportional to the slope, \(-T\,\partial y/\partial x\) for a string of tension \(T\). So the slope at the end must be zero at all times:
\[ \dfrac{\partial y_{\text{total}}}{\partial x}(\text{end}, t) = \dfrac{\partial y_{\text{incident}}}{\partial x} + \dfrac{\partial y_{\text{reflected}}}{\partial x} = 0 \]
To enforce this rule, the system generates a reflected wave whose slope at the boundary is the exact opposite of the incident wave's slope:
\[ \dfrac{\partial y_{\text{reflected}}}{\partial x}(\text{end}, t) = -\,\dfrac{\partial y_{\text{incident}}}{\partial x}(\text{end}, t) \]
A non-inverted, upright wave travelling in the opposite direction has exactly this property. So the reflected wave is an upright copy of the incident wave, and there is no phase change.
\[ y_{\text{reflected}}(x, t) = +\,f(2L - x - vt) \]
Why. mirror the incident shape about the end, and keep it upright. As before, it is a wave travelling towards \(-x\).
\[ \dfrac{\partial y_{\text{reflected}}}{\partial x} = -\,f'(2L - x - vt) \]
Why. the chain rule: the inner function \(2L - x - vt\) has derivative \(-1\) with respect to \(x\).
\[ \dfrac{\partial y_{\text{reflected}}}{\partial x}(L, t) = -\,f'(L - vt) = -\,\dfrac{\partial y_{\text{incident}}}{\partial x}(L, t) \]
Why. at \(x = L\), and since the incident slope is \(\partial f(x - vt)/\partial x = f'(x - vt)\). This is the free-end rule.
\[ y_{\text{string}} = f(x - vt) + f(2L - x - vt) \]
Read it. superpose the two waves. This is the string of the next figure on \(x \le L\).
Before you press play, predict how high the ring will rise.
An upright pulse of height \(A\) runs along a string to a free end, a ring on a smooth rod. How high does the ring rise as the pulse reflects?
In the figure above the pulse is again \(0.8\,\text{cm}\) high, starts at \(x = 3\,\text{m}\) and runs at \(2\,\text{m/s}\). At \(3.5\,\text{s}\) the readout shows the ring at \(1.60\,\text{cm}\), twice the pulse height. Then an upright pulse of \(0.8\,\text{cm}\) runs back.
A beautiful consequence: the double amplitude
The zero-slope condition has a dramatic effect on the motion of the end of the string. The relation from the chapter Introduction to Waves, \(v_{\text{particle}} = -(\text{slope}) \times v_{\text{wave}}\), explains it. For the incident wave the wave velocity is \(+v\). For the reflected wave, travelling the other way, it is \(-v\).
\[ v_{p,\,\text{incident}} = -(\text{slope}_i) \times (+v) \]
Why. the incident wave's push on the end: its velocity is \(+v\), towards the end.
\[ v_{p,\,\text{reflected}} = -(\text{slope}_r) \times (-v) \]
Why. the reflected wave's push on the end: its velocity is \(-v\), away from the end.
\[ \text{slope}_r = -\,\text{slope}_i \]
Why. the free-end rule: the two slopes cancel at the end.
\[ v_{p,\,\text{reflected}} = -(-\text{slope}_i) \times (-v) = -(\text{slope}_i) \times (+v) \]
Why. put step 3 into step 2. The two minus signs from the slope and the velocity cancel.
\[ v_{p,\,\text{total}} = v_{p,\,\text{incident}} + v_{p,\,\text{reflected}} = 2\,v_{p,\,\text{incident}} \]
Read it. step 4 equals step 1, so both waves push the end in the same direction with the same speed.
Both waves push the end of the string in the same direction with the same speed. So the total velocity of the end is doubled at every instant. Since the end moves at twice the velocity, it travels twice the distance. It reaches, for a moment, an amplitude double that of the incident wave. This overshoot is not a coincidence. It is the physical form of the string satisfying its zero-slope boundary condition.
The two ends mirror each other. At a free end the slopes cancel and the velocities add, so the end swings to \(2A\). At a fixed end the displacements cancel, and the two waves push the end in opposite directions, so it never moves.
A pulse \(3\,\text{cm}\) high reaches the free end of a string, a light ring on a smooth rod. To what greatest height does the ring rise, in cm?
At some instant, the incident pulse alone would move the free end of a string upwards at \(0.5\,\text{m/s}\). How fast does the end actually move at that instant, in m/s?
A pulse travelling on a string reaches a wall where the string is rigidly fixed. What happens to the energy of the pulse?
Which statements about the reflection of a pulse are true?
The ground surface is a free boundary for an earthquake wave arriving from below: nothing above it holds it back. For a wave coming straight up, the surface moves with about twice the amplitude of the incoming wave. A free end of a string swings to \(2A\) in the same way. Seismologists allow for this doubling when they work out the size of the waves deep inside the Earth from records taken at the surface.
Now return to the slinky. With the far end clamped, the end must stay still, so the reflected pulse is inverted: it comes back on the other side. With the far end on a light thread, the end is nearly free. The slope there must stay zero, so the pulse comes back upright, on the same side. And the free end whips out to about twice the height of the pulse.
- Saying that a free end stays still.A free end keeps its slope at zero. Its displacement is free, and it swings to \(2A\).
- Expecting the reflected pulse at a fixed end to be smaller because the wall took some energy.A rigid wall does no work and takes no energy. The inverted pulse returns at full size.
- Forgetting that the reflected wave moves with velocity \(-v\) in \(v_{\text{particle}} = -(\text{slope}) \times v_{\text{wave}}\).Use \(+v\) for the incident wave and \(-v\) for the reflected wave. The signs then show that both push a free end the same way.