Question

Take Figure 9.5c as a model for a one story building. Use the parameter values m = 1.20×103kg and k = 4.00×104 N/m. Apply the initial conditions x(0) = 1.20m, ˙ x(0) = 4.00m/s and determine the system’s free vibration response analytically (i.e. solve by hand). Express the solution as a sum of sine and cosine functions and in the amplitude-phase shift form. Plot both forms using Matlab for a span of several seconds to ensure the solutions are equivalent.

2. (5 points) Take Figure 9.5c as a model for a one story building. Use the parameter values m = 1.20 x 102kg and k = 4.00 x 104 N/m. Apply the initial conditions x(0) = 1.20m, *(0) = 4.00m/s and determine the system's free vibration response analytically (i.e. solve by hand). Express the solution as a sum of sine and cosine functions and in the amplitude-phase shift form. Plot both forms using Matlab for a span of several seconds to ensure the solutions are equivalent. TIITTIITIITIT TIITTIITIIN (a) (b) Figure 9.5 Approximating a single- story building as spring-mass system

Answer 1

Solution

Free Body Diagram

Using Newton's law,

-Kx =mx mät kx= 0

This is the equation of motion for the given system.

a) Sum of sine and cosine function

X= Asin wt tBcos wt i = Aw cos ot-Bw sinwt i- -Aw sinot -Bucoscot %3D - wm (Asinut tBoswt)+k (Arineut +B cos cut)=o - wmtk= 0 4x104 5.7735 radys 1.2x103 alt)= Asin (S-7735t)+ Bos (5.7735t) X(0) = A Sin (s-7735(0))+Bus (5.7735 () o+B 1.2= B= 1-2 i(0) = A(5.7735) Cos (5.7735c0))-1.2(5-7735)Ssin (5.7735(0)) 4 = A (5:7735) %3D A = 0.6928 0.6928 sin (5:7735t) +(-2 Cos (5-7735t) x(t)=

Plot of displacement for 5 seconds

1.5 0.5 -0.5 -1 -1.5 0. 4 3 Time, t (in seconds) 2 x(t), in m

b) Amplitude - phase shift form

x= X cos (est + 4) i: - Xw Sin(utt$) - mw X cos lutt¢)+kX Cos (ott)= 0 muttk=o 4xl04 l-2XL03 = 5:7735 vads alt)= X cos (5-7735t +¢) a10) = XCOS (S-7735 ()+4) il0)= -Xw sin(ut td) 4--X(5-T135) sin (5-7735t+ $) 4= -X(5:7735) Sin(5-7735 (0)+0) 4- -x(5:T135) sinp Dividing © bg O, tinp= -4 1.2X S.7135 $= -30° X= 1-2 cos (9) z t)= 1:3856 cas (5.1135t - 30) = 1:3856m

Plot of displacement for t=5 s

1.5 0.5 -0.5 -1 -1.5 0. 1 3 2 Time, t (in seconds) 4 X(t), in m

Comparing both the plots the solutions for the system are equivalent.