What Type of Wood is My Violin Made From 1

What type of wood is my violin made from? I have been playing the violin for 9 years, and over the course of that time have replaced my instrument three times. My current violin, the best quality of the three, is significantly louder when playing compared with those of my peers. I was intrigued to find out why this is the case, as people my parents, friends, and even my teachers - comment on it, and it’s something I have never been able to explain to them with more than a shrug and a confused smile. One reason for the loudness of the violin could be the type of the wood used for the soundboard (see Figure One) of the instrument. The soundboard is made of softwood, Different most forms often of spruce 1. spruce, with different densities, can affect sound quality and volume - for example, white spruce2, with a density of 0.431g/cm3, and Sitka spruce3, with a density of 0.45g/cm3. Meanwhile, the back is usually made of hardwood like maple, poplar, pear or willow4 due to their stiff and sturdy properties desirable in a frame, and their visual effects. The fingerboard, pegs, tailpiece and endpin are traditionally made of dense woods like mahogany or ebony. Figure One: an annotated diagram of my violin5. However, my violin is made entirely of the same wood, with exceptions of the fingerboard, tailpiece and strings. This could affect the sound quality in comparison with a traditional violin, for example, if it were entirely tonewood. A violin’s loudness is determined by the vibrations made by moving the bow across the strings, which causes the strings to vibrate. These vibrations are then transmitted through the bridge and the sound post (not shown on Figure One as it is internal) to the body of the violin. Spruce is generally used for the body as it is light and thin enough to vibrate in sympathy with the strings, acting as a sound box to radiate vibrations into surrounding air - creating a Figure Two: dimensions of my violin louder sound. The quality of wood affects this resonance and ability to vibrate, so I chose to investigate this. Other factors affecting tone quality and loudness are the tension and material of the strings, the placement of the sound post and the bow quality. Density in g/cm 3 can be found by dividing the weight of the wood by the volume. However, there are different elements to the violin (Figure One) which must be taken into account. First, the violin is weighed (without the detachable chinrest). Then, the weights of other non-removable parts must be subtracted. The volume of parts made of ebony 6 - the four pegs, the endpin and the fingerboard - can be modelled using the instrument’s dimensions (Figure Two). By multiplying this and ebony’s density (in g/cm 3), the weight of these items can be found and subsequently subtracted from the overall violin weight. Next, the weight of additional non-ebony parts must be found - the weights of the metal strings, the plastic and metal tailpiece and the wooden bridge were available on Amazon in the product detail section. Subtracting these from the weight of the violin (after subtracting ebony parts) provides the final weight of the desired wood. The volume of the violin’s wooden parts (of the species being investigated) can be attained by modelling the violin’s curved shape, explained in more depth below, using integration to find its area. An assumption must be made to estimate the hollow inner part and subtract that from the violin’s volume by assuming a uniform wood thickness of 0.2cm. The neck, pegbox and scroll of the violin (made of the same type of wood) should be taken into account as well, modelled like the ebony parts using the violin’s dimensions (Figure Two). The weight of those specific wooden parts can then be divided by their total volume of to find the wood density. Finally, this can be compared with values for wood density for specific species, and thus the species can be identified. Modelling the Violin I began by taking a photo of my violin from a birds-eye-view angle and uploading it to the Geogebra graphing tool on the internet 7. I decided to apply basic graphs to my photo which was superimposed on the graph. The image (Figure Three) shows that each small square representing 1cm2, and the scale of both x and y axis is 1 square = 1cm. Graph One: Geogebra screen capture with image of violin on graph Graph Two8 Graph Three9 Graph Four10 From here, I experimented with the shapes of graphs I knew: for example, f(x) = √ax (Graph Two) and f(x) = asin(bx + c) (Graph Three). I had covered these two graphs in class, and I could manipulate their shapes by changing the value of a, b, c and d to fit the graphs to the violin’s curved shape as accurately as possible. In addition to this, I used quadratic graphs (Graph Four) which we also studied in class, and I applied graph transformations we had learnt to alter the graphs to fit around the violin’s outline. Using the formula y = a(bx + c) 2 + d, I changed the values of a, b, c and d to fit continuously around the violin’s outline. I only did this for the outline above the x axis, because, as the violin is symmetrical, I would simply be able to double the area I find beneath the graphs, when I complete the integration. Although most of the plotting was done using trial and error, I used Geogebra to place points - E, I and K - on the maximum (and for I, the minimum) points on the violin’s body. These turning-point coordinates were then into the incorporated equations I was creating. With Geogebra, I found the points at which each of the 6 graphs I plotted Graph Five: Geogebra screen capture with violin and graphs modelling it intersect - labelled as G, H, J, L and M on the graph below (Figure Four). I used these as limits when integrating each function to find limits when integrating each function to find the area under the entire cumulative shape of point D to C (made up of sections of the 6 graphs). However, I rounded the number to 4 significant figures. This made the numbers more manageable to repeatedly input into my calculator for integration, which made my calculations more efficient and more likely to be correct. Despite the fact that some precision was lost, working to 4sf provides enough accuracy for my investigation. I also used the x intercepts of the entire violin curve - D at (0,0) and C at (36.2,0) - as limits, as this was the total length of the violin body (Figure Two). Integration of the Functions I applied the limits I had found from the intersections to the functions I had graphed (Graph Five). The equations ready for integration were as follows: f(x) = √17x for (0< x< 2.501) f(x) = 8.494sin(0.15x+0.5) for (2.501< x< 9.227) y = 0.06(x-11.94)2 + 7.639 for (9.227< x< 11.75) y = 0.08(x-17.02)2 + 5.420 for (11.75< x< 23.12) y = -0.5(0.4x-11.2)2 + 10.31 for (23.12< x< 34.57) y = -(x-33.5)2 + 8 for (34.57< x< 36.32) I chose f(x) = 8.494sin(0.15x+0.5) for (2.501< x< 9.227) to integrate by hand, and doublecheck the answer on my calculator. This is shown below (to 4sf): By hand: Using TI-84 Plus calculator: Because I found the same answer, I was then able to continue with the calculator for the rest of the calculations as follows (all to 4sf): f(x) = √17x for (0< x< 2.501) = y = 0.06(x-11.94)2 + 7.639 for (9.227< x< 11.75) = The sum of the answers listed above will give the total area under the curve of the violin - in other words, half of the area of the soundboard of the violin (not rounded to 4sf yet): 10.87 + 53.74 + 19.67 + 71.58 + 107.39 + 6.933 = 270.183cm2 Therefore, this number, when multiplied by 2, gives the area of the entire soundboard: 278.193 x 2 = 540.366cm2 The volume of the violin body can be found when the area of the soundboard is multiplied by the depth of the instrument (Figure Two) (to 5sf because the difference between 4 and 5sf can be considered significant here, as it is 4mm 3): 540.366 x 3.85 = 2080.4cm3 Integration of the Translated Functions This, however, could not give the final answer, as the violin is a hollow instrument. Because I couldn’t photograph the interior of the violin, I could not model the inside the same way as I had done the outside. Therefore, I assumed a uniform wood thickness of 0.2cm and translated all the functions down by 0.2cm (Graph Six): Graph Six: Geogebra screen capture with new translated functions f(x) = √17x - 0.2 for (0< x< 2.501) These functions then must be integrated again, in order to find the area underneath each curve between the limits of the original curve (new points of intersection were not worked out after the first round of integration). The calculations are as follows: f(x) = √17x - 0.2 for (0< x< 2.501) = f(x) = 8.494sin(0.15x+0.5) - 0.2 for (2.501< x< 9.227) = The sum of these answers will give the area of half of the interior face of the violin (not visible from outside): 10.37 + 52.40 + 19.18 + 69.31 + 105.1 + 6.580 = 262.94cm2 This number multiplied by 2 gives the total area of the interior of the face: 262.94 x 2 = 525.88cm2 This area multiplied by the depth will give the interior area (which is, for the purpose of this investigation, completely filled with air). However, the depth of 3.85cm must be reduced by two lots of 0.2cm, as the wooden structure is assumed to have this uniform thickness. Therefore, the depth of the violin body is 3.45cm: 525.87 x 3.45 = 1814.3cm3 Because the interior is hollow, the interior volume subtracted from the exterior volume will result in the volume of the wooden structure that is the violin’s body: 2080.4cm3 - 1814.3cm3 = 266.1cm3 Additional Wood Parts to Model The violin’s neck and head (made up of the pegbox and the scroll) (Figure One) are also made from the same wood as the body. Therefore, their volume must be calculated in order to find an accurate measure of the density. The neck can be measured using the diagram to the right (Figure Three). 11 19.3 cm m It can be viewed as half a cylinder. Where r = radius and h = height, the volume of a cylinder11 is: V = π r2 h By inputting the values (shown in Figure Three), the volume can be found: Figure Three: diagram of neck of violin 2.7 ÷ 2 = radius = 1.35cm π(1.35)2 = 5.726cm2 5.726 x 19.3cm = 110.51cm3 To find half of the cylinder’s volume: 110.51 ÷ 2 = 55.26cm3 The head is made up of a Bas e= 3.4c m Heigh t= 2.2cm 0.4c m pegbox and a box, as show in the picture to the right (Figure Four). My attempts to model these items are also shown to the right, using two cones to model the scroll (Figure Five) and a pair of connected parallelograms to connect them (Figure Six). I decided to use two cones to model the scroll because it is an Figure Four: scroll and pegbox of the violin unusual shape, incorporating a sort of logarithmic spiral - this would be very 2.4c m 7.8c m 2.3c m Figure Five: model of the scroll Figure Six: model of the pegbox difficult to model and would also likely be inaccurate. The cones are approximately the same volume as the scroll would be, if the spirals were evened out. The formula for the volume of a cone 12 is: V = (⅓)πr2(h) I then input the dimensions (Figure Five) into the formula above, to find the volume of the imagined conical shape: (⅓)π(3.4 ÷ 2)2(2.2) = (⅓)π(1.7)2(2.2) = 6.658cm3 I then doubled this value to find the volume of both cones (when placed back-to-back they resemble the spiral-like scroll): 6.658 x 2 = 13.32cm3 The pegbox was easier to model, with two parallelogram prisms and a cuboid as the base to connect them, as modelled above (Figure Six). To find the volume of the prisms, the area of the parallelograms13 first had to be calculated using the formula: Area = b x h = 7.8 x 2.4 = 18.72cm2 The volume of the prism could then be found by multiplying this area by the depth (Figure Six). This was then multiplied by two to account for both sides of the pegbox: 18.72 x 0.4 = 7.488cm3 7.488 x 2 = 14.98cm3 The area of the base could be found by measuring the entire width of the pegbox and removing the depth of both parallelogram prisms, then multiplying it by the length (Figure Six). Next, using an estimated wood thickness of 0.4cm (based on the prisms’ depths), the area is multiplied by the depth to give the volume: 2.3 - 2(0.4) = 2.3 - 0.8 = 1.5cm 1.5 x 7.8 = 11.7cm2 11.7 x 0.4 = 4.68cm3 The total volume for the pegbox, adding together the parallelogram prism sides and cuboid base, is: 14.98 + 4.68 = 19.66cm3 Therefore, the sum of all the non-ebony wooden parts of the violin (the body, the neck, the scroll and the pegbox) is: 266.1 + 55.26 + (13.32 + 19.66) = 354.34cm3