Answer:
Position of mass at any time (t) is given as;
U(t) = (8/11)( Cos 2t - Cos 6.928t)
Explanation:
W = mg
And from the question, W = 4lb
And g= 32 ft/s²
So m = W/g = 4/32 = 0.125
We are given that; F(t) = 8cos(2t) lb and k=6lb/in
Thus, we can describe this as;
mu'' + ku = F(t)
Ans so;
0.125u'' + 6u = 8cos(2t) - - - - eq(1)
From the question, u'(0) = 0;
And u(0) = 0.
Thus;
0.125u'' + 6u = 0
Now in simple form, this can be written as;
0.125p² + 6 = 0
So p² = -6/0.125 = -48
Since negative;
p = ±√48i
So p = ±6.928i
Since the roots are a pair of complex conjugates, the solution of eq 1 is;
Uc(t) = C1Cos 6.928t + C2Sin 6.928t
Now, the particular solution of eq 1 is given as;
Y(t) = ACos 2t + Bsin 2t
Where the coefficients A and B are determined by transforming eq 1 by putting Y(t)
Thus; eq 1 becomes ;
0.125Y''t + 6Y(t) = 8cos(2t)
And so;
0.125Y''(t) + 6Y(t) = ACos 2t + Bsin 2t
So,
0.125(-4Acos2t - 4Bsin2t) + 6(2Acos2t + 2Bsin2t) = 8cos 2t
So;
-0.5ACos 2t - 0.5BSin2t + 12ACos2t + 12Bsin 2t = 8Cos 2t
So;
11.5A Cos 2t + 11.5B Sin 2t = 8cos 2t
By ratio, A = 8/11 while B=0
And so the particular solution is;
Y(t) = (8/11)Cos 2t
The general solution of equation 1 is now;
U(t) = Uc(t) + Y(t) = C1Cos 6.928t + C2Sin 6.928t + (8/11)Cos 2t
From initial conditions, U(0) = 0amd so;
0 = C1 Cos 6.928(0) + C2 Sin 0 + 8/11 Cos0
Cos 0 = 1 and thus
C1 + 8/11 = 0 and so C1 = - 8/11
Now also, u'(0) = 0 and so;
U' = -6. 928C1 Sin 6.928t + 6.928C2 Cos 6.928t - 6.928 Sin 2t
So at u'(0) = 0;
0 = -6. 928C1 Sin 0 + 6.928C2 Cos 0 - 6.928 Sin 0
6.928 C2 = 0 and thus C2 = 0
So the position of the mass at anytime t is;
U(t) = (-8/11)Cos 6.928t + (8/11)Cos 2t
Or U(t) = (8/11)( Cos 2t - Cos 6.928t)
