Good morning! Can someone help me please? Thanks!

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1 Answer
Jul 30, 2018

Motion ( t s, h cm ) : Nadir ( 0, 0 ) ( 0.52, 1 ) ( 0.92, 3 ) ( 1.22, 5 ) ( 3, 14 ) Zenith.Motion ( t s, h cm ) : Zenith (3, 14 ) Nadir( 3.52, 13 ) ( 3.92, 11 ) ( 4.22, 9 ) ( 6, 0 ) Zenith

Explanation:

Set t and h = 0, at the time of rest at nadir. Then

h=7cos((π3)(t+3)+7=7cos((π3)t+π)+7

=7cos((π3)t)+7[0,14]cm, with h = 0, at t = 0.

h7=7cos((π3)t)

The period =2ππ3=6s

Inversely,

t=(3π)arccos(1h7)sec.

See graph, for the inverse this vertical oscillation, by ( t sec, h cm )

plots. Slide graph to see more.
graph{(x-(3/pi)arccos( 1 - y/7 ))( y^2-1)(y^2-3^2)(y^2-5^2)=0[0 6 -8 8]}

The required Table:

( i ) ( t, 1 ) ): ( (3/pi arccos(6/7), 1 )

= ( (3/3.1416)(0.541rad ), 1 )

=(0.52s,1cm).

( ii ) ( t, 3 ) ): ( (3/pi arccos(4/7), 3 )

= ( (3/3.1416)(0.963 rad ), 3 )

=(0.92s,3cm).

( iii ) ( t, 5 ) ): ( (3/pi arccos(2/7), 5 )

= ( (3/3.1416)(1.281 rad ), 3 )

=(1.22s,3cm).

Reverse motion is similar. See answer.