Wind sculpture. Cable 2. Cable 1. Sculpture

Transcription

1 Win sculpture Your frien, an artist, has been thinking about an interesting way to isplay a new win sculpture she has just create. In orer to create an aural as well as visual effect, she woul like to use the wires to hang the sculpture as some sort of string instrument. She ecies to set it up as shown in the iagram below. The sculpture is to hang from a wall an a roof overhang. Her basic esign involves two pieces of steel wire from two eye-hooks on the wall an roof, an then hanging the 5.5-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 istance from both the roof an the wall ( in the figure), an that istance shoul be about one meter. The aural effect that she woul like to achieve is that when the win blows across the two cables, the horizontal cable plays a C 3 (30.8 Hz) an the other one plays a higher note so they form a perfect fifth (i.e., the ratio of the frequencies of the two souns is :3). Your frien intens to buy the wire from a piano string retailer. She brought you the technical information she got from their webpage (next page). She nees your help figuring out the wire she nees to buy an the etails about the geometry of the installation. She only wants to buy one spool of wire, so you nee to use the same one for both segments. Cable Cable Sculpture

2 PIANO WIRE AMWG Diameter Linear ensity Resistance (mm) (kg/m) (kg) 4/ / /

3 Solution General line of reasoning: ) A cable uner tension prouces the soun that correspons to the first normal moe. ) The wavelength of staning wave epens on the length of the cable. 3) The frequency of the wave epens on the wavelength an the wave spee. 4) The wave spee epens on the tension in the wire an the mass per unit length of the wire. 5) To fin the tension, we nee to use Newton s secon law in the system for a static case (a = 0) Setup an evaluation For a cable with both ens fixe, the first normal moe correspons to this staning wave profile: Therefore, if the length of the cable is L, the wavelength of the wave is λ = L The frequency of that oscillation is relate to the wavelength through the wave spee v: v f An the wave spee epens on both the tension in the cable T (not to be confuse with perio, please) an its mass per unit length, so the frequency of the first moe is: T f L [Equation ] In this problem, we have two ifferent cables of ifferent length. The length of the cable epens on the chosen angle between the cable an the vertical irection: L L cos [Equations ] m

4 On the other han, the free boy iagram at the sculpture hook allows us to write two equations base on Newton s secon law: Horizontal: T sin T 0 (since a 0) hor Vertical: T cos mg 0 (since a 0) Therefore, T mgtan mg [Equations 3] T cos ver T T mg If we use equations an 3 in equation, we can obtain expressions for the frequencies of the first moe for each of the cables, as a function an of, the mass per unit length of the wire: mgtan f cos mg mgcos f cos [Equations 4] We know that the ratio of these two frequencies has to be :3 for a perfect fifth. If we take the ratio of the two lines in equations 4, the cancellations leave us with a expression for the angle only: f f sin cos Since the ratio between the frequencies is a number, let us call it n while we solve the equation (in our case, n = /3) sin n [Equation 5] cos Take the square of both sies of equation 5:

5 sin n cos We can write the cosine in terms of sine, since sin cos : sin n sin An that can be rearrange to a stanar quaratic equation for sin : n sin sin n 0 This has two solutions: For n = /3, this yiels sin.338 thus: 4n sin n 4 9. We only want the positive solution, so sin 0.380, an 8.3 If is fixe (.0 m), the angle ictates all the geometry (we will o that in the conclusions part below.) Once the angle has been etermine, we will use the first equation in 4 to etermine the require linear mass ensity of the wire: mgtan f For =.0 m, m = 5.5 kg, f = 30.8 Hz, =.3, we obtain a require ensity of = kg/m. Conclusions Your frien nees a wire with = kg/m. If she wants to buy it from a piano wire retailer, the closest she can get is a #8, with = kg/m. Since the table inclues the resistance of the wires, we can check if this installation is safe. Using our numerical values in equations 3, we obtain the tensions in the wires: T 03 N T 70 N

6 The quote resistance for a #8 wire is 53 kg, which correspons to 50 N. Our values are safely below this maximum. The geometric specifications are then as follows: L.00 m L.08 m.3 Or, you coul work out all the trigonometry an tell your frien exactly where she nees to install the hooks on the wall an beam. You coul give her, for instance, the istances D an D to the corner where the wall an beam meet (see figure): D D.00 m D tan.4 m D Hook Hook m For extra precision: Since the require value of oes not match one of the commercially available wires exactly, the souns will be a little off. If the artist wants to be really precise, she might want to consier changing the value of a little. Using the linear mass ensity of a #8 wire in the first line of equations 4, we can solve for the require value of : 5.5 kg9.8 m/s tan.3 mgtan m f 30.8 s kg/m With this value of, she will be able to use stanar #8 piano wire an obtain a precise C3. The values of L, L, D an D will change accoringly.