What is the velocity of the simple harmonic motion problem?

Position of the graph is
y= .4*cos(2pi*1/3*x)+1.2

I need to find the velocity as well as the acceleration. I have worked through it and keep getting that the velocity is 0. I don't believe that is correct.

You do not have time in your equation, only a solid unmoving cosine function.

Do you mean t instead of x ?

y = .4 cos[ (2 pi/3) t ] + 1.2 ?

if so then

dy/dt = v = -.4(2 pi/3) sin[ (2 pi/3) t ]

a = d^2y/dt^2 = -.4(2 pi/3)^2 cos[ (2 pi/3) t ]
which is
a = - (2pi/3)^2 * y

To find the velocity and acceleration in a simple harmonic motion problem, you need to differentiate the equation for position with respect to time. In this case, the equation is:

y = 0.4 * cos(2π/3 * x) + 1.2

Since we are dealing with time, we need to express x in terms of time (t). The relationship between x and t usually involves the equation:

x = ω * t

Where ω represents the angular frequency. In this case, we can see that the coefficient of x in the equation is (2π/3), which is equal to ω. Therefore, we can substitute x with (2π/3 * t) to rewrite the equation for position:

y = 0.4 * cos(ω * t) + 1.2

Now, we can differentiate this equation with respect to time to find the velocity:

dy/dt = -0.4 * ω * sin(ω * t)

By substituting ω with (2π/3), we get:

dy/dt = -0.4 * (2π/3) * sin((2π/3) * t)

Simplifying this expression, we have:

dy/dt = -0.8π/3 * sin((2π/3) * t)

Now, we can see that the velocity equation is not zero. It depends on the value of t. To find the velocity at a specific time, substitute that time into the velocity equation.

Similarly, to find the acceleration, we differentiate the velocity equation with respect to time:

d^2y/dt^2 = -0.8π/3 * ω * cos(ω * t)

Substituting ω with (2π/3), we get:

d^2y/dt^2 = -0.8π/3 * (2π/3) * cos((2π/3) * t)

Simplifying this expression, we have:

d^2y/dt^2 = -1.6π^2/9 * cos((2π/3) * t)

Again, we can see that the acceleration is not zero and depends on the value of t. To find the acceleration at a specific time, substitute that time into the acceleration equation.

Remember to check your calculations and make sure you have correctly differentiated the position equation.