Wind Sculpture (Involves statics, frequency)

Question

Your friend, an artist, has been thinking about an interesting way to display a new wind sculpture she has just created. In order to create an aural as well as visual effect, she would like to use the wires to hang the sculpture as some sort of string instrument. She decides to set it up as shown in the diagram below. The sculpture is to hang from a wall and a roof overhang. Her basic design involves two pieces of steel wire from two eye‐hooks on the wall and roof, and then hanging the 31.0‐kg sculpture in such a way that one of the cables is perfectly horizontal. For visual reasons, the point where both cables snatch onto the sculpture hook must be at the same distance from both the roof and the wall (d in the figure), and that distance should be about one meter. 

The aural effect that she would like to achieve is that when the wind blows across the two cables, the horizontal cable plays a C3 (130.8 Hz) and the other one plays a higher note so they form a perfect fifth (i.e., the ratio of the frequencies of the two sounds is 2:3). So, I need to find the linear mass density of the wire that could produce the frequency ratio: however, the wire must be the same type for both the horizontal part and the other part attached to the roof.

Answer 1

Hi Celia D.,

Your artist friend is no doubt familiar with the fundamental equation of a vibrating string, namely, f = (T/(m/L))^0.5 / 2L , even if you aren't(?). And probably she realizes, even if you don't(?), that the angle the roof-cable makes, is the critical thing. So, since the roof-cable will be supporting the entire 31 kg sculpture, let's say T(ang)cosθ = mg is her first step, where θ is the angle to the vertical, and T(ang) is the tension in the angled wire. That will carry as baggage, as it were, T(ang)sinθ of lateral force; which is balanced out exactly by the horizontal attachment wire (T(hor)).

Now you have to get fancy: the two frequencies can be set up as equations, where the horizontal L is 1 m, and the angled L is a function of the angle geometry (in a right triangle).

Rather that try to carry cos^-1(mg/T(ang)) through though, it's probably best to call cosθ = x, then sinθ = (1-x²)^0.5 .

Then L(ang) is 1m * (1+tan²(θ))^0.5 . Work that out as f(x); then set up the two frequency equations and their relational equation. I think you now have three equations, in three unknowns (x, m, and L). Solve!

FYI, your friend will presently discover that bare strings this short are frightfully hard to excite with gentle breezes. She may either add an inconspicuous ram-scoop, or perhaps solar-recharged-battery harmonic assist drivers, such as are used on museum-scale Foucault pendulums (my recommendation!).

-- Cheers, -- Mr. d.