Answer: y(t)= 1/π^2 sin(6*π^2*t)
Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:
y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.
The data are: ν= 3π; y(t=0)=0 and y'(0)=6.
Firstly we know that 2πν=ω then ω=6*π^2
Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0
Besides y'(t)=6*π^2*A*cos (6*π^2*t)
y'(0)=6=6*π^2*A*cos (6*π^2*0)
6=6*π^2*A then A= 1/π^2
Finally the equation is:
y(t)= 1/π^2 sin(6*π^2*t)
