Boundary Value Problem based on specific problem

3 visualizaciones (últimos 30 días)
Parvesh Deepan
Parvesh Deepan el 28 de Feb. de 2024
Comentada: Torsten el 29 de Feb. de 2024
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?

Respuesta aceptada

Torsten
Torsten el 28 de Feb. de 2024
Movida: Torsten el 28 de Feb. de 2024
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 comentarios
Parvesh Deepan
Parvesh Deepan el 29 de Feb. de 2024
sorry my bad.
The BCs are x(0) = 0.02 & x'(0) = 0.25
Torsten
Torsten el 29 de Feb. de 2024
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])

Iniciar sesión para comentar.

Más respuestas (0)

Productos


Versión

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by