A 24.00 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=0.3500 m, then lets it go. The mass undergoes simple harmonic motion with a period of 4.80000 s. What is the position of the mass 3.98 s after the mass is released?

I've calculated angular velocity using the equation:
w = 2pi*f
w = 2pi/period
w = 2pi/4.8s

I've tried using the following equation to calculate position:
x = Acos(wt)
x = (0.35m)*cos[(2pi/4.8s)(3.98s)]
x = 0.3486m

However this is not the correct answer. Could anybody find where I went wrong?

horizontal frictionless air track... why are you using Pi?

T=2pai rootover 1 by g

To find the correct position of the mass 3.98 s after it is released, we need to use the equation for simple harmonic motion, taking into account the phase angle.

You correctly calculated the angular velocity using the equation:
ω = 2π / T
where T is the period of oscillation. In this case, ω = 2π / 4.80000 s.

Now, to determine the position of the mass at a given time t, we use the equation for simple harmonic motion:

x = A * cos(ωt + φ)

In this equation, A represents the amplitude, ω is the angular velocity, t is the time at which we want to find the position, and φ is the phase angle.

To find the phase angle φ, we need additional information. The problem states that the mass is pushed to the position x = 0.3500 m and then released. From this information, we can deduce the initial conditions of the system.

At time t = 0, when the mass is released, it is at its maximum displacement, so we can set:
x(0) = A * cos(φ) = 0.3500 m

To solve for A, we can use the equilibrium position, x = 0, and substitute it back into the equation:
0 = A * cos(φ)

Since cos(φ) cannot be zero, this implies that A must be zero. Therefore, the amplitude A is zero, and the position equation simplifies to:

x = 0 * cos(ωt + φ) = 0

This means that the mass stays at the equilibrium position and does not move.

In conclusion, the position of the mass 3.98 s after it is released is at the equilibrium position x = 0.