Trigonometry help?

Question

The displacement from equilibrium of an oscillating weight suspended by a spring is given by y(t)=3sin(pit/4), "y" is the displacement and "t" is the time in seconds. 
After how many seconds the displacement will be maximum?

Answer 1

The function is y(t) = 3 sin(πt / 4).

y^'(t) = (3π / 4) cos(πt / 4)

y^"(t) = - (3π² / 16) sin(πt / 4)

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------> minimum point.

If f " (x) < 0 (negative) ------> maximum point.

To find the key numbers, y^'(t) =  0.

(3π / 4) cos(πt / 4) = 0

cos(πt / 4) = 0

cos(πt / 4) = cos (π / 2)

General solution : If cos(θ) = cos (α) then θ = 2nπ ± α.

If α = π / 2 then πt / 4 = 2nπ ± π / 2 ⇒ t = 8n ± 2.

The general form of key numbers is t = 8n ± 2.

Find the key numbers in the interval (0, 2π].

cos(πt / 4) = 0

πt / 4 = π / 2 ⇒ t = 2 and πt / 4 = 3π / 2 ⇒ t = 6.

The key numbers are t = 2 and t = 6 in the interval (0, 2π].

So, lets plug each critical point in y^"(t) = - (3π² / 16) sin(πt / 4).

If t = 2 then y^"(2) = - (3π² / 16) sin[π(2) / 4] = - (3π² / 16) < 0 (negative), therefore local maximum point.

If t = 6 then y^"(6) = - (3π² / 16) sin[π(6) / 4] = (3π² / 16) > 0 (positive), therefore local minimum point.

The displacement will be maximum after 2 seconds in the interval (0, 2π].