Posted by Sarah on .
A weight attached to a long spring is bouncing up and down. As it bounces, its distance from the floor varies periodically with time. You start a stopwatch. When the stopwatch reads 0.3 seconds, the weight reaches its first high point 65 cm above the ground. The next low point, 25 cm above the ground, occurs at 1.8 seconds. What are the next two times the spring is at a high of 50 cm after 2.0 seconds?

trig 
Reiny,
let's fit the data to a curve of the type
f(t) = a cos k(t +d) + c
from highest to lowest = 6525= 40 cm
so a = 20
from highest to lowest = .3 s to 1.8 s
so period = 3 seconds
2π/k = 3
k = 2π/3
so far we have
f(t) = 20 cos (2π/3) (t +d) + c
let's raise the curve 45 cm
f(t) = 20 cos (2π/3) (t +d) + 45
and finally , our max now occurs at t = 0, but we want to move our curve .3 to the right, so the max occurs at t=.3
f(t) = 20 cos (2π/3) (t  .3) + 45
so when does f(t) = 50 ?
20 cos (2π/3)(t  .3) + 45 = 50
cos (2π/3)(t  .3) = .25
so (2π/3)(t.3) = 1.318116
t  .3 = .62935..
t = .92935...
or (2π/3)(t.3) = 2π  1.318116
t= 2.6706....
of course since the period is 3 seconds, adding or subtracting 3 seconds from any solution will yield a new solution
so how about .92935 + 3 = 3.92935 seconds ??
So the next two times after 2 seconds to reach a height of 50 cm is
t = 2.6706 and t = 3.92935
check:
f(2.6706...) = 50
f(3.92935...) = 50 , used my calculator.