.
The function is F(t) = .
The given function describes periodic oscillations of the mass around its averaged position at 10 inches from the ceiling
with the amplitude of 6 inches and the period of 3 seconds.
(a) At the moment t= 0, when the mass was released, its distance from the ceiling was 6 + 10 = 16 inches. ANSWER
(b) At its closest position from the ceiling, the distance of the mass from the ceiling is 10 - 6 = 4 inches. ANSWER
(c) First time the weight will be at 13 inches from the ceiling at the time moment t, when
= 13, or
= 13 - 10 = 3, or
= = .
It means that at this moment
= ;
hence,
t = of a second. ANSWER
(d) The other time moments t <= 6 seconds, when the mass will be at the distance of 13 inches from the ceiling
are the OTHER solutions of the equation
= 13, or
= 13 - 10 = 3, or
= = .
So, they are
t = 3 - = 2.5 seconds;
t = 3 + = 3.5 seconds, and
t = 6 - = 5.5 seconds.
All questions are answered -- the problem is solved.