in essence it's just asking on getting the equation of that line in more or less slope-intercept form, hmmm well, it's a line, so all we need is two points to get it, hmmm let's see let's use the starting point and terminal point, so we know it goes through the origin, (0,0) and also goes through (1, 0.7), let's change 0.7 to a fraction, so its 7/10, ok, let's use those two points from it.
I only need #16 this Test/homework is hard. Help
1 answer:
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The given displacement at time, t, is
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.