An object is in simple harmonic motion. find an equation for the motion given that the frequency is 3⁄π and at time t = 0, y = 0, and y' = 6. what is the equation of motion?

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 results is a decrease in air temperature with an increase of altitude.

explanation:

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length