A merry-go-round takes 15 seconds to complete one revolution/spin.

Question

Within that time, each horse moves up and down five times. The vertical
motion of the horse spans a range of 50 cm, and the horse is 1 m high at its
vertically centre position.
a. Sketch a graph of the height of the horse over time for one complete
revolution/spin of the ride, starting with the horse at its lowest
position, where h is the height of the horse in centimetres and t is
the time elapsed in seconds.
b. State the equation of the SINE function that relates the height of the
horse, h, as a function of time, t.

Answer 1

h = 100 - a sin (2 pi t/T - pi/2)
-pi/2 so min at t = 0 where sin (-pi/2) = -sin pi/2 = -1
a = 50cm/2 = 25
T = 15/5 = 3 seconds
h = 100 - 25 sin (2 pi t/3 - pi/2)
sketch for t = 0 to t = 15

Answer 2

typo
h = 100 + a sin (2 pi t/T - pi/2)
h = 100 + 25 sin (2 pi t/3 - pi/2)

Answer 3

Our equation should have the form:
h = a sin k(t - Ø) + d

since the range of the up/down (ud) is 50 cm
a = 25
It dows 5 ud's in 15 seconds, so we have a period of 1 ud/5 sec
then 2π/k= 3 ---> k = 2π/3

so far we have h = 25 sin (2π/3 t) + 100
we also have to move our curve 3/4 units to the right, so

h = 25sin ((2π/3)(t - 3/4)) + 100

check: when t=0 , h = 25sin(2π/3(-3/4)) + 100 = 75 cm , correct, lowest as asked for
when t= 3/4 , h = 25sin(0) + 100 = 100, correct
when t = 6/4, h = 25sin((2π/3)(6/4-3/4))+100 = 125 , correct
when t = 9/4, h = 25sin((2π/3)(9/4 - 3/4))+100 = 100, as expected