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The height of a swing above ground can be modelled by equation : h=1.5+sin 3t when is...
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To know when the swiing is at 1 mt from the ground you have to solve eaquation :
That means: `sin 3t= 1-3/2=-1/2`
This causes: `3t= 11/6 pi+2kpi` `t=11/18 pi +2/3 kpi`
and: `3t=7/6 pi+2kpi` `t=7/18 pi +2/3 kpi`
So solution are `t_1=1/18 pi (11+12k)` `t_2=1/18 pi (7+12k)`
Let you see, red line is function y= 3/2+ sin3t, while blue line is function y=1
The point you are searching for are the point wich red line intercept blue one.
TO BE CONTINUE.....
Posted by oldnick on June 3, 2013 at 2:06 PM (Answer #1)
The height can be modelled by the following equation.
When the swing is 1m above ground h = 1m
`1 = 1.5+sin3t`
`-0.5 = sin3t`
`-sin(pi/6) = sin3t`
We know that
`sin(pi+theta) = -sintheta `
Let us say;
`sin(pi+pi/6) = sin3t`
`3t = (7pi/6)`
`t = (7pi)/18`
`t = 1^0 13^0 18''`
Answer to the nearest tenth second;
`t = 1^0 13^0 20''`
The calculations are considered as motion starting by increment angle.
Posted by jeew-m on June 3, 2013 at 2:09 PM (Answer #2)
Note that the funcytion has the same period T of function y= sin 3t
Indeed: if `3/2+sin3t_1=3/2+sin3t_2 rArr sin 3t_1= sin3t_2`
that is: `t_2= pi-t_1` I one round of circle.
Posted by oldnick on June 3, 2013 at 2:25 PM (Answer #3)
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