Boundary Value Problem based on specific problem

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Parvesh Deepan
Parvesh Deepan 2024-2-28
评论： Torsten 2024-2-29
Ran in:
clear all;
close all;
clc;
%% INPUTs:
f = 6; % Natural Cyclic Frequency (1/sec or Hertz-Hz)
x0 = 0.02; % Initial Displacement (in m)
v0 = 0.25; % Initial Velocity (in m/s)
%% OUTPUTs:
wn = 2*pi()*f % Natural Circular Frequency or Angular Frequency (rad/s)
wn = 37.6991
T = 1/f % Fundamental Time-Period (sec)
T = 0.1667
A = sqrt((x0^2) + (v0/wn)^2) % Amplitude (m)
A = 0.0211
vm = A*wn % Maximum Velocity (m/s)
vm = 0.7943
am = vm*wn % Maximum Acceleration (m/s/s)
am = 29.9462
Phi = atand(x0*wn/v0) % Phase Angle (in degree)
Phi = 71.6559
syms X(t)
E = diff(X,t,2) + (wn^2)*X == 0;
x = dsolve(E) % C1 & C2 are constant and can be determined by BCs
x = 
%% I need to find constant C1 & C2 through boundary value problem as x(0) = 0 & x'(0)=0. Can someone help me out?

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采纳的回答

Torsten
Torsten 2024-2-28
移动：Torsten 2024-2-28
x(0) = 0 gives C1 = 0, x'(0) = 0 gives C2 = 0. Thus the solution of your equation is x = 0 for all t.
  2 个评论
Torsten
Torsten 2024-2-29
Ran in:
clear all;
close all;
clc;
%% INPUTs:
f = 6; % Natural Cyclic Frequency (1/sec or Hertz-Hz)
x0 = 0.02; % Initial Displacement (in m)
v0 = 0.25; % Initial Velocity (in m/s)
%% OUTPUTs:
wn = 2*pi()*f % Natural Circular Frequency or Angular Frequency (rad/s)
wn = 37.6991
T = 1/f % Fundamental Time-Period (sec)
T = 0.1667
A = sqrt((x0^2) + (v0/wn)^2) % Amplitude (m)
A = 0.0211
vm = A*wn % Maximum Velocity (m/s)
vm = 0.7943
am = vm*wn % Maximum Acceleration (m/s/s)
am = 29.9462
Phi = atand(x0*wn/v0) % Phase Angle (in degree)
Phi = 71.6559
syms X(t)
E = diff(X,t,2) + (wn^2)*X == 0;
dX = diff(X,t);
conds =[X(0)==0.02,dX(0)==0.25];
x = dsolve(E,conds) % C1 & C2 are constant and can be determined by BCs
x = 
fplot(x,[0 1])

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