A spring is attached to the ceiling and pulled 9 cm down from equilibrium and released. The amplitude decreases by 7% each second. The spring oscillates 15 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Respuesta :

Answer:

[tex]D(t) = 9(0.93)^t cos(30 \pi t)[/tex]

Explanation:

Amplitude begins at 9 cm, [tex]A_0 = 9 cm[/tex]

The amplitude decreases by 7% (0.07) each second

The amplitude function can then be modeled as:

[tex]A(t) = A_0(1 - 0.07)^t\\A(t) = 9(0.93)^t[/tex]

The spring oscillates 15 times each second, the period of oscillation (time to make 1 oscillation) will therefore be calculated as:

T = 1/15

[tex]\frac{2\pi }{B} = \frac{1}{15} \\\\B = 30\pi[/tex]

The graphical equation of the system described is:

[tex]D(t) = A cos ( Bt - C) + D[/tex]

Horizontal shift, C = 0

Vertical shift, D = 0

[tex]D(t) = 9(0.93)^t cos(30 \pi t)[/tex]