Wind chimes Part I

Fig.1

There is a lot of information online about wind chimes. However, I was unable to find an answer to a natural question: is there a critical wind speed below which the wind chimes will not emit any sound?

The key to answering this question is to first find out the response of a pendulum to a constant horizontal force. This is solved as a question at the website:

http://www.physics.utah.edu/~woolf/2210_ailion/exam3_11.pdf

“A ball of mass m is connected by a strong string of length L to a pivot. A steady wind exerts a constant horizontal force F on the ball as shown. Initially, the ball is held in place with the string vertical. The ball is then released and is observed to swing up and down between it staring point and its maximum height H above its starting point.” - by David Ailion of the University of Utah.

Ailion gives the worked out solution, but does not specifically connect it to the wind-speed v. This is expressed as follows:

                                F = rv²A/2

where r is the density of air and A is the area of the wind-catching surface of the pendulum bob. This assumes that the angle between the normal to the surface and the wind velocity f = 0. If not, this force would be reduced by a cos(f) factor.

                Although Ailion has given the worked-out solution for the equilibrium height and for the maximum height, this is mine; the final equations are the same.

However, the geometry is somewhat different for wind chimes – there is a pendulum bob which catches the wind, but there is also a striker, which is much further up, and is a thin round disk which actually contacts one of the hollow cylindrical chimes, producing the tone that we hear. This geometry is dealt with in the second section, and then applied to the cases of bamboo and iron wind chimes.

Pendulum with constant horizontal force F:

see Fig.1:

a)      Equilibrium height: velocity = maximum

h = L [1 – cos(q)]

Resolve tension T into vertical and horizontal components:

                                T sin(q) = F

                                T cos(q) = mg

                                tan(q) = F/mg

Substituting into equation for h:

h = L{ 1 – [ 1 + (F/mg)²]^-1}

If F = 0, the equilibrium position of the pendulum is the standard h = 0 (pendulum hangs vertically down when there is no wind).

If F = mg, h = L/√2,

And if F = ¥, h = L.

Note: H.Munera & H.R.Maya Lat.Am.J.Phys.Ed. 5 (2011) 22-30: 'method of apparent vertical' applied to the "perturbation of a dynamical system by a constant or nearly constant force, not necessarily small. The basic idea is to rotate the system of coordinates so that the z-axis coincides with an apparent acceleration of gravity".

(Munera & Maya claim that their method is not the same as the equivalence principle. I was unable to understand why, and it is anyway a finer point that is not relevant here). This gives a vector equation:

g' = g + F/m

The upshot is that the equilibrium position is displaced by an angle q, given by: tan(q) =  F/mg. This gives the same position as derived above, as it must.

b) maximum height: velocity = 0

For a constant horizontal force acting over a horizontal distance x:

                                Fx = mgh

This x is related to the length of the pendulum by:

                                x = L sin(q)

The height h is given by:

                                h = L[ 1 – cos(q)]

Eliminating h we get:

F/mg = [1 – cos(q)]/sin(q) = tan (q/2)

Substituting in the equation for h and doing some algebra, using the half-angle formula:

                                tan(q) = 2 tan(q/2)/[1 – tan²(q/2)]

We finally get the maximum height:

                                h = 2L/[1 + (mg/F)²]

If F = mg, h = L.

If F = ¥,  h = 2L.

Fig.2:

Fig.2 shows the wind blowing with velocity v from L to R, displacing the equilibrium position from the vertical dotted line to the line marked 'eq'. The position 1 corresponds to the maximum height calculated above. The position 2 corresponds to the maximum height on the other side as the pendulum swings.

The equilibrium position is a measure of the energy due to the wind because h_eq = 0 when F = 0.

Thus the height corresponding to 2 is given by:

                                mg(h₁ – h₂) = mgh_eq

When there is no wind h_eq = 0 and h₁ = h₂ and the pendulum swings symmetrically about the equilibrium position.

Actually, some of the analysis is not needed for the stated purpose: to find the critical wind speed at which the wind chimes will sound. The only one needed is the one for the maximum height:

                                h = 2L/[1 + (mg/F)²]

However, the actual geometry of the wind chime is somewhat more complicated than a simple pendulum in a wind. There is a bob (which catches the wind) and there is a striker which is higher up, which actually touches the hollow metal cylinders that actually emit the sound on being struck. The actual case is discussed in the second section.

Light & Height

OK, I’m not anywhere near being Flaubert, so I do not seek the mot juste.

I’m also declaring my independence to blog about the nebulous and quite possibly the trite and the obvious – and to do it shamelessly.

The Sun is largely high in the sky, ignoring sunrises, sunsets and high latitudes.

We tend to replicate this in our homes, although lights are more often on walls than ceilings – and rarely on floors.

Trees (almost) always grow upwards, towards the big light in the sky. They are phototropic.

Height is an advantage – the taller trees have unfettered access to sunlight.

For humans, the taller you are, the farther you see. In the savannah that humans are widely supposed to have grown up in, height would allow one to spot both predators and prey in advance. Of course, one could have climbed up a tree to scan the far horizon – if there happened to be a tree nearby.

In the sea also, a tall captain would help – but sailors solved the problem of spotting storms early by putting a lookout on the mast.

Anyway, the area that can be scanned is directly proportional to the height – of the human or the vantage point.

But the point is that the upward direction is associated with more light and increased viewing area. Conversely, if you are shorter or if you are stuck in a pit or a deep valley, you will have a smaller amount of light and a decreased viewing area. So a downward direction is associated with less light.

An extreme example is the concept of building a very high tower at the South Pole that is permanently in sunshine - or a solar collector in low Earth orbit. Either could be used to generate solar energy.

The tower would be 527 kms high on Earth, assuming a perfectly spherical Earth (for R=6400 and q=22.5°). Of course, you would have to add about 18 kms to the height of the tower, since the polar radius (6353 kms) is less than the mean radius (6371 kms). 

Note: The geometry is obtainable from the gif link mentioned below.

http://en.wikipedia.org/wiki/Earth_radius

http://theworld.com/~reinhold/lunarpolar.html