If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1 lb/in is suddenly set in motion at t = 0 by an external force of 2 cos 7t lb, determine the position of the mass at any time. (Use g = 32 ft/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in feet of the mass from its equilibrium position at time t seconds.)

U(t) = (48/45)(cos 7t - cos 8t) seconds

Explanation:

Weight = 6lb

K = 1lb/in or 12lb/ft

U'(0) = 0 ft

U''(0) = 0 ft/s

F(t) = 2 cos 7t lb

We know that, w = mg

Thus, m = w/g = 6/32 = 3/16 lb²/ft

Thus, the initial value problem cam be written as;

3u'' + 192u = 48 cos 7t

Since, u'(0) = 0 ft and u''(0) = 0 ft/s, the corresponding homogenous equation is;

3u'' + 192u = 0

With a characteristic of ;

3p² + 192p = 0

So, p = ±√(192/3)i = ±√64i = ±8i

This is a pair of conjugate roots and thus the solution is;

u(t) = c1cos 8t + c2sin8t

And the particular solution can be written as;

Y(t) = A cos 7t + B sin 7t

Now, let's plug Y(t) into the initial value problem;

3Y''(t) + 192Y(t) = 2 cos 7t

3(-49Acos 7t - 49Bsin 7t) + 192(A cos 7t + B sin 7t) = 48 cos 7t

Thus;

-147ACos 7t - 147BSin 7t + 192A cos 7t + 192B sin 7t = 48 cos 7t ;

45A Cos 7t + 45B Sin 7t = 48 cos 7t

By inspection,A = 48/45 and B=0

Hence, the particular solution is;

Y(t) = (48/45)cos 7t

The general solution will now be;

U(t) = Uc(t) + Y(t)

U(t) = c1 cos8t + c2 sin8t + (48/45)cos 7t

Mow, let's find c1 and c2 from the conditions in the initial value problem.

Thus ;

At u(0) =0;

0 = c1 cos 0 + c2 sin0 + (48/45)cos 0

So, c1 = 48/45

Also, at u'(0) = 0;

0 = - 8c1 sin 0 + 8c2 cos 0 - (7)(48/45)sin 0

8c2 = 0 and c2 = 0

Thus, the general solution of u(t) is;

U(t) = (48/45)( cos 7t - cos 8t) seconds

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80cm/s.

24cm/s.

53.3cm/s.