Solving for the Initial value conditions
- In solving for the
initial value conditions, we had the general ODE and the length relation
- Our conditions were based on the axis we chose which was ranging between (-a/2...a/2)
- The conditions are in more detail in the following picture especially in number 2 and 3:
- In our first experiment we chose the end points to be at 32 cm apart and the density of the chain to be 3.96 g/cm and we got that (c1 = 0), to solve for D and the other integrating factor
we used maple for that::
- To solve for D:
fsolve(1267/x = sinh(633.6/x), x);
D = -291.0373088, but if we look at the equation again we notice that negative and positive values for D
give the same answer because of the sin properties. So which value to choose? We chose the positive value because that stands for the Least potenial graph, while if we choose the negative value we get the highest
potential graph.
- To solve for the other integrating factor lamda:
fsolve(x = (291)*cosh(39.6*32/(2*(291))), x);
Soln = 1300.183809
- As soon as we get the 2 integrating factors, we plot the curve and we get the following:
- As we observe from the curve, the least potential is at ( 0, -0.26)
- Also we observe that the curve cuts the x-axis in two points, (-.16,0) and (0.16,0) which makes sense
because the curve is from -0.32 to 0.32
- Now the next step was that we took sample points on the curve and compared them to our catenary curve we built:
(-1300/(3960)) + (291 * cosh(3960*(-.1)/(291))/(3960));
-.1755865612 (-.1,-.17)
(-1300/(3960)) + (291 * cosh(3960*(.1)/(291))/(3960));
-.1755865612 (.1,-.17)
(-1300/(3960)) + (291 * cosh(3960*(.05)/(291))/(3960));
-.2371211998 (-.05, -.237)
- We plotted those points inaddition to the x and y-intercepts on a big graph paper and we were able to compare
what we have on it to the sample curve we built and the points were very close................."Result was good"
- Our conditions were based on the axis we chose which was ranging between (-a/2...a/2)
- The conditions are in more detail in the following picture especially in number 2 and 3:
- In our first experiment we chose the end points to be at 32 cm apart and the density of the chain to be 3.96 g/cm and we got that (c1 = 0), to solve for D and the other integrating factor
we used maple for that::
- To solve for D:
fsolve(1267/x = sinh(633.6/x), x);
D = -291.0373088, but if we look at the equation again we notice that negative and positive values for D
give the same answer because of the sin properties. So which value to choose? We chose the positive value because that stands for the Least potenial graph, while if we choose the negative value we get the highest
potential graph.
- To solve for the other integrating factor lamda:
fsolve(x = (291)*cosh(39.6*32/(2*(291))), x);
Soln = 1300.183809
- As soon as we get the 2 integrating factors, we plot the curve and we get the following:
- As we observe from the curve, the least potential is at ( 0, -0.26)
- Also we observe that the curve cuts the x-axis in two points, (-.16,0) and (0.16,0) which makes sense
because the curve is from -0.32 to 0.32
- Now the next step was that we took sample points on the curve and compared them to our catenary curve we built:
(-1300/(3960)) + (291 * cosh(3960*(-.1)/(291))/(3960));
-.1755865612 (-.1,-.17)
(-1300/(3960)) + (291 * cosh(3960*(.1)/(291))/(3960));
-.1755865612 (.1,-.17)
(-1300/(3960)) + (291 * cosh(3960*(.05)/(291))/(3960));
-.2371211998 (-.05, -.237)
- We plotted those points inaddition to the x and y-intercepts on a big graph paper and we were able to compare
what we have on it to the sample curve we built and the points were very close................."Result was good"