# What is the answer for question 6) ? http://postimg.org/image/qfwflx4j1/ (Reminder): It is 1 question.

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The rest position for the spring is .5; the minimum height is .1m and the maximum height is .9m

The period is given as 1.2 seconds. The start position is at the minimum.

The general form for a sinusoid is `y=acos(b(x-h))+k` . (Here we choose cosine since the function begins at a max or min -- you can choose to use sine with a phase shift if you want.)

a gives the amplitude (and if a<0 the graph is reflected over the horizontal axis),b yields the period (`b=(2pi)/p` ), h is the horizontal translation or phase shift and k is the midline.

Here the amplitude is .4m, and a=-.4. (Began at minimum.)

`b=(2pi)/1.2=(5pi)/3`

h=0 there is no phase shift.

k=.5

So `H(t)=-.4cos((5pi)/3x)+.5`

The graph:

Then h(.3)=.5, h(.7)=.846,h(2.2)=.3

We assume that the spring has the same displacement in both directions and that the oscillations are not dampened (e.g. from friction, air resistance, etc...).