# Confused regarding the wavelength of standing waves and their relation with wave speed

Suppose we have an air column closed at one end and open at one end. We know, the general formula for the wavelength of the stationary waves of different harmonics will be equal to $$4L/n$$. My confusion is, in this equation there is no speed of wave involved. Why not though? How does this make sense? If wave speed increases won't wavelength increase (assuming frequency is kept constant)?

According to the formula above, speed will have no effect on the wavelength of the stationary wave, which is obviously not true. What am I missing here?

• What do you mean by the speed of a standing wave? Mar 24 at 9:45
• @MariusLadegårdMeyer A standing wave can be decomposed into two waves travelling in opposite directions. The wave speed of a medium still affects a standing wave, even though the phase velocity is zero. Mar 27 at 3:23
• The claim that you say "is obviously not true" is, in fact, true. Mar 27 at 3:54

. . . . . speed will have no effect on the wavelength of the stationary wave . . . . .

The wavelength of a stationary wave in a tube is determined only by the length of the tube.
The conditions are that there must be a displacement node (pressure antinode) at the closed end and a displacement antinode (pressure node) at the open end.

However, $$\rm wavelength = \dfrac {\text{speed of wave}}{\text{frequency of wave}}$$, thus the wavelength having been fixed results in the combination of speed and frequency being fixed.
For example, if the temperature of the gas in the tube increases which in turn means that the speed of the wave increases, the frequency of the standing wave also increases.

This sort of change has a material effect on the tuning of a pipe organ.
Suppose an organ has been tuned at a temperature of $$21^\circ \rm C$$ to produce a particular note when a certain key is depressed, eg a frequency of $$440\,\rm Hz$$, then if the temperature of the air increases the note produced by the organ when that same key is depressed will be of a frequency higher than $$440\,\rm Hz$$ resulting in the organ sounding off tune. The length of the organ pipe and hence the wavelength of the standing wave has not changed but the frequency of the note emitted by the organ pipe has.
The article Temperature and the Pipe Organ: A Practical Guide to Achieving Stability illustrates this idea.

I suggest reviewing standing waves, e.g. from a good school book.

You have two effects:

1. propagation of (sound) waves in a medium
2. interference, resulting in standing waves in your case

Ad #1: For a given medium, like air, water, steel, a guitar string etc., waves have the well known relationship $$c = \lambda \cdot f$$, $$c$$ speed of propagation (e.g. after exciting the medium in one point), $$\lambda$$ wavelength, $$f$$ frequency.

This is completely dependent on the medium: it's its property.

Ad #2: This is completely a geometrical effect. When you excite, e.g. a string or an air column in some point, this happens:

• in 1D, like a string or air column, two waves start propagating, say to the left and to the right or: into all available directions, which are just 2 here
• in 2D a circular wave starts propagating into all planar directions
• in 3D a radial wave starts propagating into all directions of space

From reflections you convert them into proapagating waves, heading towards each other. If geometry matches $$\lambda$$ in the right way(s), interference result in a wave, which seems to stand still. Else, it won't seem to stand still.

"If wave speed increases won't wavelength increase (assuming frequency is kept constant)?"

The wave speed, $$v$$ (that is the speed of the progressive waves of which the standing wave may be considered a superposition), is a constant for a given gas (e.g. air) at a given temperature. So the usual way to think about stationary waves in air columns (closed at one end) is that the wavelength, $$\lambda$$, is determined by the pipe length, $$L$$, according to $$\lambda=4L/n$$, as you have said, and this also determines the natural frequency, because $$f=v/\lambda$$.

It's not easy to change $$v$$ significantly: for example if the air temperature is increased from 20°C to 30°C, $$v$$ increases by only 1.7 %. Nonetheless, the frequencies produced by the pipe would rise a little as a consequence – by 1.7 % in this case! This assumes that the wavelengths produced are unaffected – true to a very good approximation, though if the pipe's temperature is raised, $$L$$, and hence the possible values of $$\lambda$$, will increase very slightly.

I believe your confusion stems from the fact that you take the given relation - $$\lambda_{n} = \frac{4 L }{n}$$ - as the one that completely determines the parameters of the standing wave. The given relation is a necessary relation that must hold for all standing waves for an open-closed tube (or some other 1D system with the same boundary conditions - like strings strings free at one end - ). The more general equation \ref{1} shown below is one that describes the system in a more complete way.

$$f_{n} = \frac{\left(2 n + 1 \right) c}{4 L}, ~~~~~~~~~~~~~~~~~ n = 1, 2, \ldots \tag{1} \label{1}$$

with $$f_{n}$$ the frequency of the standing wave, $$n$$ denoting the index of the harmonic series, $$c$$ the speed of sound and $$L$$ the length of the tube. Please note that the following equation is exactly the same with $$n$$ belonging to a different set (the odd positive numbers).

$$f_{n} = \frac{n c}{4 L}, ~~~~~~~~~~~~~~~~~ n = 1, 3, 5, \ldots \tag{2} \label{2}$$

Now, here it is rather obvious than when the speed of sound changes, the frequency of the standing wave also changes. This is to accommodate the change that must happen for the formation of a standing wave, which of course is given by the equation you provided (for odd $$n$$).

What is possibly confusing for someone here is that the relationship that must hold is independent of the speed of sound, like you mention. Nevertheless, this equation does not show "all the truth". This equation is a necessary condition that must be satisfied. If the speed of sound changes, then under those new conditions the same equation must be satisfied. For this to happen the wavelength must have a constant relation to the length of the tube.

Now, since the wavelength must be a constant value for the formation of a standing wave we could look at the relation connecting the wavelength and the speed of sound to see what is going on.

$$c = \lambda f \tag{3} \label{3}$$

As you see, when the left hand side changes, so must the right hand side of the equation in order for the equality to hold. Since we have constrained the wavelength to be constant (necessary condition for the formation of the standing wave) what is left free to change to accommodate the necessary changes is the frequency. Thus, when the speed of sound changes, the frequency will also change to one that has the same wavelength as the previous one under the new conditions.

This shows that the relation between wavelength and the length of the tube will indeed remain constant and is independent of the speed of sound.