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