A spring is attached to the ceiling and pulled 16 cm down from equilibrium and released. After 4 seconds

Question

A spring is attached to the ceiling and pulled 16 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 7 cm. The spring oscillates 6 times each second. Assume that the amplitude is decreasing exponentially. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

D(t) = 

Answer 1

In general,

D(t) = A(t)sin(2πft),

where

A(t) is the amplitude,

f = 6s^-1 is the frequency.

We are told that the amplitude is a general exponential

A(t) = A(0)e^kt

for some constant k, which we need to calculate.

We are given

A(0) = 16cm,

A(4) = 7cm

From that information we can calculate k by plugging given value into the formula for A(4)

7 = 16e^4k

k = ln(7/16)/4 = -0.2

So the answer is

D(t) = 16e^-0.2tsin(12πt)