A merry-go-round takes 15 seconds to complete one revolution/spin. 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
To Solve for the equation I did the following:
Form of equation should be h = a sin k(t - d) + c
1m height of horse = 100 cm vertical displacement = c
Range is up/down 50 cm, hence a = 50/2 = 25
Period is 5 times up/down in 15 sec, therefore 15/5 = 3. Then 2π/k = 3. Solving for k= 2π/3
d - I am unsure
So far my equation is h = 25 sin ( (2π/3 (t-d)) + 100
You are on the right track.
You have the equation $h=a\sin k(t-d)+c$
You have the range $a=25$
Since the it goes $5$ times up and down in $15$ seconds, $k=\dfrac{2\pi}{3}$
We have $h=25\sin\bigg(\dfrac{2\pi}{3}t\bigg)+100$
But note that we need to move the curve $\dfrac34$ units in the right direction, so we get $$h=25\sin\bigg(\bigg(\dfrac{2\pi}{3}\bigg)\bigg(t-\dfrac{3}{4}\bigg)\bigg)+100$$