At a construction site, a long concrete drain pipe lies on the ground, waiting to be laid. A child shouts into one end. When the far end is blocked by a heap of sand, the shout comes back as a hollow echo. When the far end is open to the air, a return still comes back, although there is no wall there at all.
Two rules from the string
The chapter Superposition and Reflection found that a boundary makes a new wave. The new wave, the reflection, is whatever keeps the boundary's rule. For a string there were two rules. A fixed end holds the displacement at zero, \(y = 0\), so the pulse returns inverted. A free end holds the slope at zero, \(\partial y/\partial x = 0\), so the pulse returns upright.
Sound in a pipe has two descriptions, from the chapter Propagation of Sound Waves. One is the displacement wave \(s(x, t)\) of the air particles along the pipe. The other is the pressure variation \(\Delta P\). They are tied together by \[ \Delta P = -B\,\dfrac{\partial s}{\partial x}, \] where \(B\) is the bulk modulus of the air. So a rule about \(\Delta P\) is also a rule about the slope of \(s\). This one link turns the two string rules into the two rules for the ends of a pipe.
Case 1: a rigid wall
Let a sound wave travel along a pipe towards a rigid wall that closes its far end at \(x = L\). The air next to the wall cannot move into the wall. It cannot leave the wall either, or a gap would open. So the displacement of the air at the wall is zero at every instant: \[ s(L, t) = 0. \] This is the fixed-end rule of the string, written for air. A rigid wall is therefore a displacement node.
The fixed-end rule gives the fixed-end answer. Let the incident displacement wave be \(s_i = f(x - vt)\). The reflected wave is the inverted mirror image, \(s_r = -f(2L - x - vt)\), exactly as on the string. So the displacement wave returns inverted, with a phase change of \(\pi\). What happens to the pressure is less obvious, and it comes from \(\Delta P = -B\,\partial s/\partial x\).
\[ s_i = f(x - vt), \qquad s_r = -\,f(2L - x - vt) \]
Why. the incident wave, and the reflected wave that keeps \(s = 0\) at \(x = L\) (the fixed-end rule of the string).
\[ \Delta P_i = -B\,\dfrac{\partial s_i}{\partial x} = -B\,f'(x - vt) \]
Why. the pressure of the incident wave, from \(\Delta P = -B\,\partial s/\partial x\).
\[ \dfrac{\partial s_r}{\partial x} = +\,f'(2L - x - vt) \]
Why. the chain rule: the inner function \(2L - x - vt\) has derivative \(-1\) with respect to \(x\), which cancels the minus sign in front.
\[ \Delta P_r = -B\,f'(2L - x - vt) \]
Why. multiply step 3 by \(-B\).
\[ \Delta P_r(L, t) = -B\,f'(L - vt) = +\,\Delta P_i(L, t) \]
Read it. put \(x = L\) in steps 4 and 2. At the wall the reflected pressure equals the incident pressure, with the same sign.
So the pressure wave returns upright, with no phase change. A compression that reaches the wall comes back as a compression. At the wall itself the two pressures add, and the pressure variation there is twice that of the incident wave. The wall is a pressure antinode. The air piles up against the wall, and nothing lets it escape.
Before you press play, predict what comes back from the wall.
A sound pulse runs along a pipe towards its closed end. Its front is a compression C. What runs back from the wall?
In the figure above the pulse starts at \(x = 3\,\text{m}\) and runs at \(2\,\text{m/s}\) to the end at \(10\,\text{m}\). These are the numbers of the string pulses, and the drawing is slowed. With the rigid wall, the readout of \(s\) at the end stays \(0.00\) the whole time. As the compression reaches the wall, near \(3.25\,\text{s}\), the readout of \(\Delta P\) climbs to \(2.00\), twice the incident peak. By \(7\,\text{s}\) an inverted displacement hump is back at \(x = 3\,\text{m}\), led by a compression.
A sound wave of pressure amplitude \(0.02\,\text{Pa}\) meets a rigid wall head on, and all of it is reflected. What is the largest pressure variation at the wall, in Pa?
Case 2: an open end
At an open end the air in the pipe meets the open atmosphere. The atmosphere is vast, and it holds the pressure at its ambient pressure \(P_0\). A small pulse of sound cannot raise or lower the pressure of the whole sky. So, to a good approximation, the pressure variation at an open end is zero at every instant: \[ \Delta P(L, t) = 0. \] An open end is a pressure node.
Now use the link. Since \(\Delta P = -B\,\partial s/\partial x\), a zero pressure variation means a zero slope of the displacement: \[ \dfrac{\partial s}{\partial x}(L, t) = 0. \] This is exactly the free-end rule of the string. So the displacement wave behaves like a string with a free end. It returns upright, \(s_r = +f(2L - x - vt)\), with no phase change. At the end the incident and reflected displacements add, and the air there swings twice as far. The open end is a displacement antinode.
The pressure goes the other way. Repeat the derivation with \(s_r = +f(2L - x - vt)\). Step 3 now gives \(\partial s_r/\partial x = -f'(2L - x - vt)\), so \(\Delta P_r = +B\,f'(2L - x - vt)\). At \(x = L\) this is \(-\Delta P_i(L, t)\). The reflected pressure is the exact opposite of the incident pressure at the end. So the pressure wave returns inverted: a compression comes back as a rarefaction.
Choose the open end in the figure above and press play again. The readout of \(\Delta P\) at the end stays \(0.00\). At \(3.5\,\text{s}\) the readout of \(s\) reaches \(2.00\): the air at the mouth of the pipe swings twice as far as the incident pulse. The pulse that runs back has an upright displacement hump and a rarefaction R at its front.
Think of a crowd pushing along a narrow corridor that opens onto a large ground. At the doorway the people spread out at once, and the crush vanishes. They rush out so freely that they overshoot, and a thin, empty region forms behind them. That thinning then travels back along the corridor. A compression reaching an open end does the same, and returns as a rarefaction.
The outside air does not hold the pressure fixed exactly at the rim of the pipe. The node of pressure lies a little beyond the open end, about \(0.6\,r\) outside a pipe of radius \(r\). Also, not all of the energy is reflected. A small part is radiated into the room, which is how we hear a flute. For wavelengths much longer than the width of the pipe, most of the energy is reflected. The chapter Standing Sound Waves uses this end correction.
A compression pulse travels along a pipe towards an open end. What travels back into the pipe?
The two ends of a pipe
The two cases give the boundary conditions for a sound wave inside a pipe. For a longitudinal wave in a pipe, the boundaries are the ends of the pipe. The behaviour of the air molecules at the ends decides where nodes and antinodes can form.
A closed end is a solid barrier, and the air molecules there are not free to move back and forth. So a closed end is always a displacement node. It is also a pressure antinode, because the air piles up against the wall.
An open end is exposed to the atmosphere, where the pressure is nearly constant. So an open end is always a pressure node. To keep the pressure constant, the air molecules there must be free to move with the largest amplitude. So an open end is also a displacement antinode.
"Inverted" or "upright" always needs a name: inverted in what? At a wall the displacement is inverted and the pressure is upright. At an open end it is the other way round. Say which wave you mean, or say "a compression returns as a compression".
The chapter Standing Sound Waves builds the notes of every wind instrument from these two rules. A flute is open at both ends, so it has a displacement antinode at each end. A clarinet is closed at one end, so it has a node at that end and an antinode at the other.
Which statements about sound at the ends of a pipe are true?
Explain why the open end of a pipe, which is a pressure node, must be a displacement antinode.
The pressure variation is \(\Delta P = -B\,\partial s/\partial x\). At an open end \(\Delta P = 0\), so the slope of the displacement is zero there. That is the free-end rule, so the reflected displacement is upright and adds to the incident displacement at the end. The air at the end swings with twice the amplitude, which makes it a displacement antinode.
Before an ultrasound scan, the doctor spreads a clear gel on the skin. Without it, a thin layer of air would lie between the probe and the skin. For a wave travelling in the probe or in the body, that air acts almost like an open end: the pressure there can hardly change. So nearly all the ultrasound would be reflected back into the probe, inverted, and almost none would enter the body. The gel removes the air, and the sound passes in.
Now return to the concrete pipe. With the far end blocked by sand, the air at the blocked end cannot move. So the shout returns as it went, a compression for a compression. With the far end open, the outside air holds the pressure there fixed. The compression spills out, overshoots, and sends a rarefaction back. So an open end reflects sound although there is nothing to hit, and its return is inverted in pressure.
- Saying that a reflected sound wave is "inverted" without saying which wave.At a wall the displacement is inverted and the pressure is upright. At an open end the pressure is inverted and the displacement is upright.
- Placing a displacement antinode at a closed end.The air cannot move into a wall, so a closed end is a displacement node. The pressure swings most there.
- Thinking that an open end lets the sound escape and reflects nothing.An open end is a pressure node. For wavelengths much longer than the pipe is wide, most of the sound is reflected, as a rarefaction for a compression.