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Suspension system of car can be modelled as mass spring damper. The equation for the position y of the mass in the mass spring damper system is second order differential equation d²y dy where m is mass of the object, b is damping ratio, k is spring constant, and f(t) is the force that is acting on the system Use central differencing to find the propagation equation and obtain the solution assuming zero condition for the position and velocity of the mass, i.e. dy y(0) 0. (0) = 0 Write Python function that given m, b, k, initial time to final time t, and time step At calculate and return y in list Use the function to plot your response for the following parameters 0 15 At 0.005 m k 5 b 0.5. 1.and 2 f(t) Your graph should show three plots, one for each value of b. Include axis labels and a legend identifying the different plots Challenge: Experiment with different values of At and determine the largest value that results in a "good approximation" of the solution for the parameters given below. You need to choose your own criteria for a "good approximation" Plot graph supporting your claim 1, k=5, b 0.5, f(t) Finite difference formulas Forward difference dx Ax Backward difference df 1-10-1-1 dx-Ax Central difference 1-4-1-1 dx 2Ax Forward difference Backward difference Central difference Euler's method Numerical solution to initial value problems (IVPs) dy f(y,t), y(0)=y dt Yi+1 =y1 Propagation Equation , At 2At 3At 44t SAt y y(0) y(At) y(2At) y(3At) y(4At) y(5At) y, y1 y2 y3 y4 11 Euler's method . Numerical solution to initial value problems (IVPs) dy dt f(y,t), y(0)= Yi+1 =y f(yw.4) Propagation Equation 5 , At 2At 3At 4At 5At y y(0) y(At) y(2At) y(3At) y(4At) y(5At) y, Yo y, y2 y3 y4 11 Euler's method . Numerical solution dt dt At Yi+1 y At t ti ti+1

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import matplotlib.pyplot as plt
import seaborn as sns
sns.set_style("whitegrid")

# y'' = (f(t) - b * y' - k * y)/m i.e.
# y'' = F(t, y, y', f)
# initial conditions:
# y(0) = 0 - initial position
# y'(0) = 0 - initial velocity

"""
*** Euler method - central differenciating ***

second derivative:
y(t)'' = F[t, y(t), y'(t), f(t)]    (1)
y(t)'' = [y(t+h) - 2y(t) + y(t-h)]/h^2    (2)

from (1) and (2) we obtain --->
[y(t+h) - 2y(t) + y(t-h)]/h^2 = F[t, y(t), y'(t), f(t)]
equation of time propagation of y:
--- > y(t+h) = 2y(t) - y(t-h) + F[t, y(t), y'(t), f(t)]*h^2   (3)

first derivative:
(y(t)')' = F[t, y(t), y'(t), f(t)]
(y(t)')' = [y(t+h)'-y(t-h)']/2h
equation of time propagation of first derivative y':
---> y(t+h)' = y(t-h) + 2h*F[t, y(t), y'(t), f(t)]    (4)
"""


def F(b, f):
    # function that returns left-hand side of our equation y'' = (f(t) - b * y' - k * y)/m
    # function has damping ratio and func f as input parameters
    # y1 = first derivative...

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