0+0+sin(0) = 0 mL L Eqn (1) We also linearised the sin (0) term to get...

  

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0+0+sin(0) = 0 mL L Eqn (1) We also linearised the sin (0) term to get the below ODE. 0 + 0 + 0 = 0. . Eqn (2) ........ mL L 1) Convert both Eqn (1) and Eqn (2) to a system of 1st order ODEs. 2) Apply ode45 to both systems of ODEs in 1) and plot the solutions on the same graph in MATLAB. Assume that c = m = L = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10. Comment on any similarities or differences. 3) Apply a for loop to vary the values of L and comment on the effect of this on the solution to Eqn (1) using ode45. Assume that c = m = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10. 4) Apply a while loop to vary the values of m and comment on the effect of this on the solution to Eqn (2) using ode45. Assume that c = L = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10. 0+0+sin(0) = 0 mL L Eqn (1) We also linearised the sin (0) term to get the below ODE. 0 + 0 + 0 = 0. . Eqn (2) ........ mL L 1) Convert both Eqn (1) and Eqn (2) to a system of 1st order ODEs. 2) Apply ode45 to both systems of ODEs in 1) and plot the solutions on the same graph in MATLAB. Assume that c = m = L = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10. Comment on any similarities or differences. 3) Apply a for loop to vary the values of L and comment on the effect of this on the solution to Eqn (1) using ode45. Assume that c = m = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10. 4) Apply a while loop to vary the values of m and comment on the effect of this on the solution to Eqn (2) using ode45. Assume that c = L = 1, g = -9.8 m/s, 0(0) = 0, 0 (0) = 10.