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

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

Explanation:

Answer 2

The motion can be modeled as

`x(t) =9 \, cos( (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

`(2 \pi t)/(T) = ( \pi )/(2)`
That is,

`t= (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( ( \pi t)/(6) )`

When t = 1 s, the position of the weight from equilibrium is

`d = 9 \, cos( ( \pi )/(6) ) = 9( (1)/(2) ) = 4.5 \, in`

Answer:

`d = 9 \, cos( ( \pi )/(6) ) = 4.5 \, in`