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.
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
typo
h = 100 + a sin (2 pi t/T - pi/2)
h = 100 + 25 sin (2 pi t/3 - pi/2)
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
a. To sketch the graph of the height of the horse over time for one complete revolution/spin, we can plot the height at different points in time.
The horse completes one full up-down cycle in 15 seconds, and within that time it moves up and down five times. So, each up-down cycle takes 15/5 = 3 seconds.
At t = 0 seconds, the horse is at its lowest position, so we start the graph from the point (0, -50) on the y-axis.
From t = 0 to t = 3 seconds, the horse completes its first up-down cycle. During this time, the height increases from -50 cm to +50 cm. Thus, we can plot points on the graph as follows: (0, -50), (1, -25), (2, 0), (3, 25), and (4, 50).
At t = 3 seconds, the horse completes one full cycle, and at t = 15 seconds, it completes one full revolution/spin. Therefore, we repeat the pattern of points at t = 0 to t = 15 seconds: (0, -50), (1, -25), (2, 0), (3, 25), (4, 50), (5, 25), (6, 0), (7, -25), (8, -50), (9, -25), (10, 0), (11, 25), (12, 50), (13, 25), (14, 0), and (15, -25).
Plotting all these points on the axes, we can then connect them with a smooth curve to represent the graph of the height of the horse over time for one complete revolution.
b. The equation of the sine function that relates the height of the horse, h, as a function of time, t, can be given as follows:
h(t) = A * sin(B * t + C) + D
In this case:
- A is the amplitude, which is half the range of motion of the horse, so A = 50/2 = 25 cm.
- B determines the frequency or the number of cycles per unit of time. Since the horse completes one full cycle in 3 seconds, we have B = 2π/3.
- C is the phase shift, which indicates the horizontal shift of the graph. Since the horse starts at its lowest position, we have C = 0.
- D is the vertical shift, which represents the average height of the horse during the motion. In this case, the horse is 1 m high at its vertically center position, so we have D = 100 cm.
Putting it all together, the equation of the SINE function to describe the height of the horse as a function of time is:
h(t) = 25 * sin((2π/3) * t) + 100