Question

A weight attached to a spring is at its lowest point, 9 inches below equilibrium, at time t = 0 seconds. When the weight it released, it oscillates and returns to its original position at t = 3 seconds. Which of the following equations models the distance, d, of the weight from its equilibrium after t seconds?

Answer 1

Answer:

D=-9cos(2pi/3t)

Step-by-step explanation:

Answer 2

The motion can be modeled as
$x(t) =9 \, cos( \frac{2 \pi t}{T} )$
where 
x = the position below the equilibrium position
T =  the period of the motion
t = time

The weight returns to its equilibrium position when x = 0. 
This occurs when
$\frac{2 \pi t}{T} = \frac{ \pi }{2}$
That is,
$t= \frac{T}{4}$

Because the weight returns to the equilibrium position when t =  3 s, therefore
T = 12 s
The motion is
$x(t) = 9 \, cos( \frac{ \pi t}{6} )$

When t = 1 s, the position of the weight from equilibrium is
$d = 9 \, cos( \frac{ \pi }{6} ) = 9( \frac{1}{2} ) = 4.5 \, in$

Answer:
$d = 9 \, cos( \frac{ \pi }{6} ) = 4.5 \, in$