Question

Good morning! Can someone help me please? Thanks!

Answer 1

Motion `uarr` ( t s, h cm ) : Nadir ( 0, 0 ) ( 0.52, 1 ) ( 0.92, 3 ) ( 1.22, 5 ) ( 3, 14 ) Zenith.Motion `darr` ( 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 = 7 cos ( ( pi/3 )( t +3 ) + 7 = 7 cos ( (pi/3) t + pi ) + 7`

`= - 7 cos ( (pi/3) t ) + 7 in [ 0, 14 ] cm`, with h = 0, at t = 0.

`rArr h - 7 = - 7 cos ( (pi/3) t )`

The period =`(2pi)/(pi/3) = 6 s`

Inversely,

`t = (3/pi) arccos( 1- h/7 ) sec`.

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

plots. Slide graph `uarr` 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 )

`= (0.52 s, 1 cm)`.

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

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

`= (0.92 s, 3 cm)`.

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

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

`= (1.22 s, 3 cm)`.

Reverse motion is similar. See answer.