SHM acceleration problem

[Punchlinegirl]

A 29.0 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=.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 3.50 s. What is the position of the mass 2.975 s after the mass is released?
I got this part fine, with an answer of .206 m
Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

I found the angular velocity by using
\omega = 2\pi/ T and got it to be 1.795. I then plugged it into the equation, a= -\omega^2 x_{m}cos(\omega t +\theta)
with x= .206 m, and t=2.975 s. I got a= .389 m/S^2, which isn't right.
can someone tell me what I'm doing wrong?

[OlderDan]

A 29.0 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=.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 3.50 s. What is the position of the mass 2.975 s after the mass is released?
I got this part fine, with an answer of .206 m
Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

I found the angular velocity by using
\omega = 2\pi/ T and got it to be 1.795. I then plugged it into the equation, a= -\omega^2 x_{m}cos(\omega t +\theta)
with x= .206 m, and t=2.975 s. I got a= .389 m/S^2, which isn't right.
can someone tell me what I'm doing wrong?

You were asked for the maximum acceleration. That is not what you calculated. Where is acceleration maximum, or what is the maximum based on your acceleration equation

[Punchlinegirl]

Ok I used the ma=-kx. Solving for k with my angular velocity gave k=93.4 N/m. Plugging in m= 29 kg and x= .206 m gave a=.664 m/s^2. This wasn't right. I understand that I wasn't solving for max acceleration before, but I don't think I know an equation to do it.

[OlderDan]

Ok I used the ma=-kx. Solving for k with my angular velocity gave k=93.4 N/m. Plugging in m= 29 kg and x= .206 m gave a=.664 m/s^2. This wasn't right. I understand that I wasn't solving for max acceleration before, but I don't think I know an equation to do it.

Look carefully at the equation that you wrote for the acceleration. Also, think about when the force is greatest and use what you know about the relationship between force and acceleration. Either of these (you don't need both, but they are connected) will get you to the answer.