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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.....
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.
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.
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