Respuesta :
A) The amplitude and period of the given motion are given as;
Amplitude = 1.803 ft ; Period = π/2 s
B) The number of complete cycles the mass will have completed at the end of 6π seconds is;
Cycles in 6π s = 12 cycles
Let M be the mass attached and let K be the spring constant.
Applying Newton's second law to this system, we have;
m(d²x/dt²) = -kx²
When x(t) is the displacement from the equilibrium position, the equation can be given as;
d²x/dt² + (k/m)x = 0 - - - - - (eq1)
Let's find the mass.
W = mg
W = 32lb and g = 32 ft/s²
So, m = W/g = 32/32 = 1 lbft/s² = 1 slug
Let's find spring stiffness;
From Hooke's law, W = ks
s = 2 ft. Thus, k = W/s = 32/2 = 16 lb/ft
Now, putting the values of m and k into eq 1,we obtain;
d²x/dt² + (16/1)x = 0
d²x/dt² + 16x = 0
Now, we know that the solution of x" + ω²x = 0 is given as;
x(t) = C1 cosωt + C2 sinωt
By inspection, ω² = 16. Thus, ω = 4
So, the general solution is now;
x(t) = C1 cos4t + C2 sin4t
The initial conditions from the question are;
x(0) = -1 ft and x'(0) = -6 ft/s
At x(0) = -1
-1 = C1 cos 0 + C2 sin 0
-1 = C1
Now, at x'(0) = 6
x' = -4C1sin4t + 4C2cos4t
So,
6 = -4C1sin 0 + 4C2cos0
4C2 = - 6
C2 = -6/4 = -3/2
Thus, the equation of motion is;
x(t) = - cos4t - (3/2)sin4t
A) Amplitude is given as;
A = √(C1² + C2²) = √(-1² + (-3/2)²)
A = √(1 + 9/4) = √(13/4) = 1.803 ft
Period is given as;
T = 2π/ω = 2π/4
T = π/2 s
B) Cycles it will complete in 6π seconds;
= 6π/T = 6π/(π/2) = 12 cycles
Read more at; https://brainly.com/question/15177368
