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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given the data:

(1) The max is about 55, the min is about 5: then the range is 50, and the amplitude is about 25, ao a=25. (`a=("max"-"min")/2` )

(2) The function goes from its max to its min in about 7 seconds, thus the period is about 14 seconds. Since `b=(2pi)/p==>b=(2pi)/14~~.449`

b is approximately .449

(3) The vertical translation gives the midline -- it is the average of the max and min so `d=(55+5)/2~~30`

(4) For the phase shift we find when the function takes on the value of the midline or 30. This happens at about t=3.5 seconds. However, at this point the function is decreasing; the sin begins at the midline and increases so we go back 1/2 a period (7 seconds).

The phase shift is 3.5 left or `c~~-3.5`

(5) The equation is close to:

`y=25sin(.449(t+3.5))+30`

* My calculator gives about 25.07sin(.46t+1.52)+29.95 using a sin regression*

** Note that there is a choice for both a and c -- we could have c=3.5 and a=-25 which will also work -- then there is any phase shift of a multiple of 14 seconds**

(6) The graph:

(7) We found the amplitude without the graph or the equation by finding `a=("max"-"min")/2`

(8) The rest position will be the functions minimum or 5m

(9) The maximum displacement is max-min=50m

(10) The time for 15 cycles is 15*14=210 seconds (3.5 minutes)

Sources:

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