# A weight is attached to a spring is pulled down 4 inches below equilibrium position. Assuming period of system is 1/3 seconds, determine a trig model that gives the position of the weight at time...

A weight is attached to a spring is pulled down 4 inches below equilibrium position. Assuming period of system is 1/3 seconds, determine a trig model that gives the position of the weight at time t seconds.

When is the best way to tell whether to solve with a sine or cosine equation. seems to get me sometimes.

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### 1 Answer

The solution is either a sine or a cosine function with time as the variable. In general the solution can be written as a sine or cosine function with an arbitrary phase(the angle C ) as:

The general formula for sine: `y = Asin(Bt - C) +D` and for cosine: `y = Acos(Bt - C) +D` where,

A- amplitude of the function = `(max-min)/2` (vertical stretch of the graph)

B - stretch/shrink on the x-axis. (It compresses or expands the graph.) B has the following relationship with the period: B=`2pi/P` , where P is the period.

`C/B` - the phase shift of the graph (the shift left (if `C/B` is neg.) or right (if `C/B` is pos.))

D - the vertical shift of the graph.

Note that the role of phase C is to identify where the object is with respect to the equilibrium position at time t=0. That is if the mass is displaced an amount A and then released at t=0, the phase is just C=0 and `y= AcosBt` . If it is passing through the origin at t=0, then C=`pi/2` and the function can be written as follows: `y=AsinBt` .

Here,the weight attached to the spring is pulled down 4 inches below equilibrium position. So, A=-4

Again, period P=`1/3` . Hence, B=`(2pi)/(1/3)=6pi `

**Therefore, the trigonometric model that gives the position of the weight at time t seconds is**: `y=-4cos6pi t . `