Question 880696: Hello! I am having difficulty with my homework understanding this question. Here is the question:
Tommy has a tree swing near the river in his backyard. The swing is a single rope hanging from a tree branch. When Tommy swings, he goes back and forth across the shore of the river. One day his mother (who was taking an adult math course) decided to model his motion using her stopwatch. She finds that, after 2 sec, Tommy is at one end of his swing, 4 m from the shoreline while over land. After 6 sec, he reaches the other end of his swing, 5.2 m from the shoreline while over the water.
b) Write the equation expressing distance from the shore versus time.
c) Predict the distance when
i) time is 6.8 sec
ii) time is 15 sec
iii) time is 30 sec
d) Where was Tommy when his mother started the watch?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We are given the period(T) of the swing, T = 2*6, now we can calculate the length (L) of the swing
T = 2*pi*square root(L/g) where g is 9.81 m/sec2
12 = 2*pi*square root(L/9.81)
square root(L/9.81) = 12 / 2*pi
square both sides of =
L/9.81 = 144 / 4*pi^2
L/9.81 = 3.644628099
L = 35.753801653 m
b) position(P) = A * sin(t), where A is the amplitude
for a small angle say 5 degrees, sin(5) = A / L
A = L * sin(5)
A = 35.753801653 * 0.087155743 = 3.116149139
now 12 seconds corresponds to 360 degrees, therefore
12/360 = 6.8/x, 12x = 360*6.8, x = 360*6.8 / 12 = 204
equilibrium position = (4 + 5.2) / 2 = 4.6 m
c.i) P = 3.116149139 * sin (204) = 1.26745204 m to the left of the equilibrium position
c.ii) P = 3.116149139 * sin (450) = 3.116149139 m to the right of the equilibrium position
c.iii) P = 3.116149139 * sin (900) = 0 m to the right of the equilibrium position
d) t = 12 -2 = 10
P = 3.116149139 * sin (300) = 2.698664316 m to the left of the equilibrium position
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