Sheet1 simplified
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The document describes a damped mass-spring system and provides the equation of motion for analyzing the free vibration of the system. It then gives the general solution to the differential equation that describes the response x(t) in terms of the system's natural frequency, damping ratio, initial displacement, and initial velocity. The student is asked to: 1. Create a Matlab function to calculate the response x(t) for given parameter values. 2. Run sample code that plots the response for different damping ratios. 3. Calculate and submit the response at two specific cases.
![49
2 Problem
Deadline 10:15 am, 29 March 2012, Submitted as Hard-copy to Dr. Ahmed Rashed
1. Consider the damped mass spring system shown below
f (t)
x (t)
m
k c (or d )
• The equation of motion can be written as
m¨(t) + cx(t) + kx(t) = f (t)
x ˙
• For free vibration, f (t) = 0 is substituted in the above equation to obtain
m¨(t) + cx(t) + kx(t) = 0
x ˙
k c √
• By defining ωn ≡ , ξ≡ & cc ≡ 2 km , the above equation
m cc
can be divided by m and written as
2
x(t) + 2ξωn x(t) + ωn x(t) = 0
¨ ˙
• The solution of the above differential equation is given to you as
A sin (ωn t + θ)
,: ξ = 0
−ξωn t
B e sin (ωd t + φ) ,: ξ < 1
x(t) = −ωn t
e
[x0 + (ωn x0 + v0 ) t]
√ √
,: ξ = 1
2 −1t 2 −1t
Ce−ωn ξ
−ξωn t
e + Deωn ξ ,: ξ > 1
where x0 is the initial displacement, v0 is the initial velocity and the terms ωd ,
A, B, C, D, θ and φ are as follows
ωd ≡ ωn 1 − ξ2
2
v0
A≡ x2 +
0
ωn
x2 ωn + 2ξωn x0 v0 + v0
0
2 2
B≡
ωn 1 − ξ2
26](https://image.slidesharecdn.com/sheet1-simplified-120325205759-phpapp01/85/Sheet1-simplified-1-320.jpg)
![ξ 2 − 1 − ξ ωn x0 − v0
C≡
2ωn ξ2 − 1
ξ 2 − 1 + ξ ωn x0 + v0
D≡
2ωn ξ2 − 1
x0 ωn
× × MISLEADING × × θ ≡ tan−1 × × MISLEADING × ×
v0
and
x0 ωn 1 − ξ 2
× × MISLEADING × × φ ≡ tan−1 × × MISLEADING × ×
ξωn x0 + v0
(1) Create a Matlab function called “SDOF_response” to calculate x(t) as follows
function x=SDOF_response(w_n,zeta,x0,v0,t)
• w_n, zeta, x0 and v0 are scalars
• t is a vector
Common mistake
• Use the “atan2” function for calculating θ and φ
• “atan2(y,x)” function uses y and x values to determine the correct
angle quadrant. Hence it returns angles in the closed interval [−π, π]
• “atan” function on the other hand returns angles in the range [− π , π ]
2 2
(2) Print and submit your “SDOF_response.m” file
(3) Study and understand and run the following code
clc
w_n=1;
zeta=[0,.1,.15,.2,.3,.5,1/sqrt(2),1,2];
legend_string={'$zeta = 0$','$zeta = 0.1$','$zeta = 0.15$','$zeta =
0.2$','$zeta = 0.3$','$zeta = 0.5$','$zeta=1/sqrt{2}$','$zeta =
1$','$zeta = 2$'};
x0=-1;
v0=0;
t=linspace(0,4*pi,500);
figure(1);clf
hold all
for n=1:length(zeta)
x=SDOF_response(w_n,zeta(n),x0,v0,t);
plot(w_n*t,x,'LineWidth',1.5)
end
grid on
title('x(t) for omega_n = 1, x_0 = -1 and v_0 = 0');
xlabel('omega_n t');
legend();
set(gca,'XTick',0:pi/2:4*pi);
set(gca,'XTickLabel',{'0','','pi','','2pi','','3pi','','4pi'})
set(gca,'XLim',[0,4*pi]);
set(gca,'YTick',-1:.5:1);
legend(legend_string,'interpreter','latex','Location','NorthEastOutside');
27](https://image.slidesharecdn.com/sheet1-simplified-120325205759-phpapp01/85/Sheet1-simplified-2-320.jpg)
Sheet1 simplified
- 1. 49 2 Problem Deadline 10:15 am, 29 March 2012, Submitted as Hard-copy to Dr. Ahmed Rashed 1. Consider the damped mass spring system shown below f (t) x (t) m k c (or d ) • The equation of motion can be written as m¨(t) + cx(t) + kx(t) = f (t) x ˙ • For free vibration, f (t) = 0 is substituted in the above equation to obtain m¨(t) + cx(t) + kx(t) = 0 x ˙ k c √ • By defining ωn ≡ , ξ≡ & cc ≡ 2 km , the above equation m cc can be divided by m and written as 2 x(t) + 2ξωn x(t) + ωn x(t) = 0 ¨ ˙ • The solution of the above differential equation is given to you as A sin (ωn t + θ) ,: ξ = 0 −ξωn t B e sin (ωd t + φ) ,: ξ < 1 x(t) = −ωn t e [x0 + (ωn x0 + v0 ) t] √ √ ,: ξ = 1 2 −1t 2 −1t Ce−ωn ξ −ξωn t e + Deωn ξ ,: ξ > 1 where x0 is the initial displacement, v0 is the initial velocity and the terms ωd , A, B, C, D, θ and φ are as follows ωd ≡ ωn 1 − ξ2 2 v0 A≡ x2 + 0 ωn x2 ωn + 2ξωn x0 v0 + v0 0 2 2 B≡ ωn 1 − ξ2 26
- 2. ξ 2 − 1 − ξ ωn x0 − v0 C≡ 2ωn ξ2 − 1 ξ 2 − 1 + ξ ωn x0 + v0 D≡ 2ωn ξ2 − 1 x0 ωn × × MISLEADING × × θ ≡ tan−1 × × MISLEADING × × v0 and x0 ωn 1 − ξ 2 × × MISLEADING × × φ ≡ tan−1 × × MISLEADING × × ξωn x0 + v0 (1) Create a Matlab function called “SDOF_response” to calculate x(t) as follows function x=SDOF_response(w_n,zeta,x0,v0,t) • w_n, zeta, x0 and v0 are scalars • t is a vector Common mistake • Use the “atan2” function for calculating θ and φ • “atan2(y,x)” function uses y and x values to determine the correct angle quadrant. Hence it returns angles in the closed interval [−π, π] • “atan” function on the other hand returns angles in the range [− π , π ] 2 2 (2) Print and submit your “SDOF_response.m” file (3) Study and understand and run the following code clc w_n=1; zeta=[0,.1,.15,.2,.3,.5,1/sqrt(2),1,2]; legend_string={'$zeta = 0$','$zeta = 0.1$','$zeta = 0.15$','$zeta = 0.2$','$zeta = 0.3$','$zeta = 0.5$','$zeta=1/sqrt{2}$','$zeta = 1$','$zeta = 2$'}; x0=-1; v0=0; t=linspace(0,4*pi,500); figure(1);clf hold all for n=1:length(zeta) x=SDOF_response(w_n,zeta(n),x0,v0,t); plot(w_n*t,x,'LineWidth',1.5) end grid on title('x(t) for omega_n = 1, x_0 = -1 and v_0 = 0'); xlabel('omega_n t'); legend(); set(gca,'XTick',0:pi/2:4*pi); set(gca,'XTickLabel',{'0','','pi','','2pi','','3pi','','4pi'}) set(gca,'XLim',[0,4*pi]); set(gca,'YTick',-1:.5:1); legend(legend_string,'interpreter','latex','Location','NorthEastOutside'); 27
- 3. (4) submit your resulting figure x(t) for ωn = 1, x0 = −1 and v0 = 0 1 ζ=0 ζ = 0.1 ζ = 0.15 0.5 ζ = 0.2 ζ = 0.3 ζ = 0.5 0 √ ζ = 1/ 2 ζ=1 ζ=2 −0.5 −1 0 pi 2pi 3pi 4pi ωn t (5) Calculate and submit the response, x(t), at the following cases (a) m = 1, c = 2, k = 3, x0 = 0.4, v0 = 1 and t = 0.5 (b) m = 2, c = 3, k = 4, x0 = 0.4, v0 = −1 and t = 0.5 28