Question

The equation of motion for a weight suspended from a particular spring is given by p(t)=2sint3+5cost3p(t)=2sin⁡t3+5cos⁡t3, where p(t)p(t) is the displacement from the equilibrium position in centimeters, and tt is the time elapsed in seconds. at what time does the weight first reach the equilibrium position?

Answer 1

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.