Question

consider the system shown where m=50kg, c=200N.s/m, k1=350N.m, and k2=550N.m. The free end of the spring k2 is excited by y(t)=0.4sin3t(m) as shown

4. Consider the system shown where m = 50 kg, c = 200 N.s/m, ki = 350 N.m, and k2 = 550 N.m. The free end of the spring ky is excited by y(t) = 0.4 sin 3t (m) as shown (20 points) a) Determine the equation of motion of the system. b) Determine the natural frequency and damping ratio of the system. c) Using ode45 of MATLAB, simulate and plot the response x(t) of the system with initial conditions x(0) = -0.1 m and +(0) = 0.2 m/s. In your simulation use tfinal = 15 s. k * , ko y(t) m HMI ch

Answer 1

Solution

* kz y(t) my 1 mm

Free Body Diagram

- དྷ ན པ

Using Newton's law,

kaly - 2) - kır – ci = më

mö + ci + (ki+k2.c = kay

mi +ci + (ki+k2.= 0.4k2 sin 3t

This is the equation of motion of the system.

* + (%)*+ (* # 2) - ()

$\omega_n^2=\frac{k_1+k_2}{m}$

Iki + kr wn = VmV . 350 + 550 - = 4.2426 rad/s 50

Therefore natural frequency of the system is 4.2426 rad/s

# %, =2(2) ཡ = ཟེར་བ་

= 2 wn

200 = = 24.2426)

ζ = 0.4714

Therefore damping ratio of the system is 0.4714.

System response

10 Time (s) (ա սլ) (1}x

MATLAB Code

function system_response

t=0:0.001:15; % time scale

%System parameters
m=50;
c=200;
k1=350;
k2=550;

%Initial conditions
initial_x = -0.1;
initial_dxdt = 0.2;

[t,x]=ode45( @f, t, [initial_x initial_dxdt] );

plot(t,x(:,1));
grid on
xlabel('Time (s)'); ylabel('x(t) (in m)');

function dxdt=f(t,x)
dxdt_1 = x(2);
dxdt_2 = -(c/m)*x(2) - ((k1+k2)/m)*x(1) + (k2/m)*0.4*sin(3*t);

dxdt=[dxdt_1; dxdt_2];
end
end