An undamped spring-mass system contains a mass that weighs 4lb and a spring with spring constant 6lb/in. It is suddenly set in motion at t=0 by an external force of 8cos(2t)lb⁡. Determine the position u of the mass at any time t. Use 32ft/s2 as the acceleration due to gravity. Pay close attention to the units.

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)

8 meters above the spring

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