Problem 2.16: The potential of a free movable point mass m in a horizontal plane is given by V(x, y) = k(x^2 + y^2)/2. a) Determine the acting forces and formulate the equations of motion. b) Determine the path of the point mass in parameter and implicit representation for the initial conditions x(0) = a, ˙x(0) = 0, y(0) = 0, ˙y(0) = v0. MATLAB: % Problem 2.16: Potential of a free movable point mass in a horizontal plane % Given parameters k = 1; % Arbitrary constant m = 1; % Mass of the point mass a = 1; % Initial x position v0 = 1; % Initial velocity in y direction % Define symbolic variables syms x(t) y(t) % Define potential function V = @(x, y) k*(x^2 + y^2)/2; % Equations of motion eqn1 = diff(V(x, y), x) == diff(diff(x, t), t) * m; eqn2 = diff(V(x, y), y) == diff(diff(y, t), t) * m; % Initial conditions x0 = a; y0 = 0; xDot0 = 0; yDot0 = v0; % Solve the differential equations [xSol(t), ySol(t)] = dsolve([eqn1, eqn2, x(0) == x0, y(0) == y0, diff(x)(0) == xDot0, diff(y)(0) == yDot0]); % Display the solutions disp(['x(t) = ' char(xSol)]); disp(['y(t) = ' char(ySol)]); Python: # Problem 2.16: Potential of a free movable point mass in a horizontal plane # Importing required libraries import sympy as sp # Given parameters k = 1 # Arbitrary constant m = 1 # Mass of the point mass a = 1 # Initial x position v0 = 1 # Initial velocity in y direction # Define symbolic variables t = sp.symbols('t') x = sp.Function('x')(t) y = sp.Function('y')(t) # Define potential function V = k*(x**2 + y**2)/2 # Equations of motion eqn1 = sp.diff(V, x) - sp.diff(sp.diff(x, t), t) * m eqn2 = sp.diff(V, y) - sp.diff(sp.diff(y, t), t) * m # Initial conditions x0 = a y0 = 0 xDot0 = 0 yDot0 = v0 # Solve the differential equations solution = sp.dsolve([eqn1, eqn2], [x, y], ics={x.subs(t, 0): x0, y.subs(t, 0): y0, x.diff(t).subs(t, 0): xDot0, y.diff(t).subs(t, 0): yDot0}) # Display the solutions print('x(t) =', solution[0]) print('y(t) =', solution[1])