Respuesta :
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)
The equation of the simple harmonic motion is; y = ¹/₆sin 6t
What is the equation of the motion?
The general expression for a harmonic motion is;
y(t) = A sin (ωt + φ)
where;
A is the amplitude.
ω is the angular frequency.
φ is phase constant
Now, we know that 2πf = ω and we are given frequency; f = 3/π
Thus; 2π * 3/π = ω
ω = 6
Thus, since at t = 0, y = 0, then;
0 = A sin(6(0) + φ)
A can't be zero and as such the value of φ has to be 0 or π.
Now, y'(t) = ω²A cos(ωt + φ)
since at t = 0, y' = 6, then we have;
6 = 6²A cos(6*0 + 0)
6 = 36A
A = 6/36
A = 1/6
Thus, the equation is; y = ¹/₆sin 6t
Read more about simple harmonic motion at; https://brainly.com/question/26114128
