Answer:
Explanation:
By the general expresion for this problem, we have:
y(t) = A*cos(w*t+∅)
since: w = 2π/T = 2π/0,65
For the initial conditions:
y(0) = (0.17cm)*cos(w*0+∅) = + 0.17 m ---> cos(∅) = 1 ---> ∅ = 0°
Then:
A) y(t) = (0.17 m)*cos((2π/0,65)*t)
<u>B part</u>
This means, find the first solution for:
y(t) = (0.17 m)*cos((2π/0,65)*t) = 0 > (equilibrium position)
then: cos((2π/0,65)*t) = 0 ---> (2π/0,65)*t = π/2 ---> t = 0.1625 sec
<u>C part</u>
By definition: (Velocity) v = dy/dt
Then, deriving: v = dy/dt = - (0.17 m)*(2π/0,65)*sin((2π/0,65)*t)
The maximum velocity ocurrs when sin((2π/0,65)*t) = ±1, then (in absolute value): Vmax = 1,64 m/s
<u>D part</u>
By definition: (Aceleration) A = dv/dt
Then, deriving: v = dv/dt = - (0.17 m)*(2π/0,65)²*cos((2π/0,65)*t)
The maximum aceleration ocurrs when cos((2π/0,65)*t) = ±1, then (in absolute value): Amax = 15,88 m/s²
<u>E part</u>
<em>For the acceleration</em>, ocurres by all the solution when:
cos((2π/0,65)*t) = cos(phase) = ±1, this means: phase = {π, 2π, 3π, ...}, all π multiples.
Then, for the position:
<em>y(t) = (0.17 m)*±1= {+0.17 m ; -0.17 m}</em>
<em>For the velocity</em>, ocurres by all the solution when:
sin((2π/0,65)*t) = sin(phase) = ±1, this means: phase = {π/2, π, 3π/2, ...}, all π/2 multiples.
Then, for the position:
<em>y(t) = (0.17 m)*cos(phase)= 0</em>
