Question

6. Undamped Vibrations: Solve the initial value problem for y(t). y" +y = cos(wt); w2 #1; y(0) = 0; y'(0) = 0. (8) Plot y(t) versus t, for w= -0.2, 0.9 and 6 to observe beats and resonance.

Answer 1

D 6. y"+y=- Los (WE) wtf 2000 initial condition yo) ro, y'co)=o. homogeneous solution : Yast) m2+1=0 ot .] mati y (t) = e (Acost tbrint) A cost + Brint - A -> yo)=0 y'10) 20 => Bao :.9 (t)=0. solution : sple) Particulas solution: wswt Sp (t)= D²+1 coswt ( -(W)? +1 cosul 1-w2 9plt)= The general solution is y(t) = Yalt)typits Y't) Coswl ابی - ا -

w=0.2 1.5 system response static deflection 1 0.5 y(t) 0 -0.5 -1 -1.5 5 10 15 25 30 35 40 20 time w=0.9 6 system response -static deflection 4 2 -2 -4 -6 0 5 10 15 25 30 35 40 20 time w=6 1 system response static deflection 0.8 0.6 0.4 0.2 WWW -0.2 0 5 10 15 25 30 35 40 20 time

clc;clear all;close all;
t=0:0.01:40;
w=0.2;
figure
y=cos(w.*t)./(1-w.^2);
plot(t,y)
xlabel('time')
ylabel('y(t)')
hold on
y1=1+0.*t;
plot(t,y1)
legend('system response','static deflection')
title('w=0.2');

w=0.9;
figure
y=cos(w.*t)./(1-w.^2);
plot(t,y)
xlabel('time')
ylabel('y(t)')
hold on
y1=1+0.*t;
plot(t,y1)
legend('system response','static deflection')
title('w=0.9');

w=6;
figure
y=cos(w.*t)./(1-w.^2);
plot(t,y)
xlabel('time')
ylabel('y(t)')
hold on
y1=1+0.*t;
plot(t,y1)
legend('system response','static deflection')
title('w=6');

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The natural frequency of the system is 1rad/sec

1)Now,when the forcing frequency is much lower than the narural frequency of the system,

Then the response of the system is aproximatly equals to the static deflection of the system due to applied force.

2)When the forcing frequency is to the narural frqeuency of the system,

Then the response of the system is going more value.

3) when the forcing frequency is much larger than the narural frqeuency of the system,

Then the response of the system is very smaller than the static deflection of the system dut to applied force.