Solving for Amplitude of a Brick on a Piston with SHM: Help Needed!

[Mayerzee]

I have no idea why I can not get this question, but here it is:

A brick is resting atop a piston that is moving vertically with simple harmonic motion of period 1.19 s. At what amplitude will the brick separate from the piston?

I know that once the acceleration is greater than that of gravity the 2 will seperate.

I tried using the equation:
a(t)=-w^2Acos(wt+phi) a=-9.8 w=2pi/T phi=pi and t=T

Where am I wrong on this question?
Any help would be greatly appreciated! Thanks
Peter

 

[Doc Al]

You are on the right track. What's the equation for y(t)?

 

[wais]

Hi Peter,

It seems like you have the right idea with using the equation a(t)=-w^2Acos(wt+phi) to solve for the amplitude of the brick on the piston. However, there are a few things that may be causing your difficulty in getting the correct answer.

First, when using the equation a(t)=-w^2Acos(wt+phi), it is important to remember that the amplitude (A) represents the maximum displacement from the equilibrium position. In this case, the equilibrium position is when the brick is resting on the piston. So, when solving for the amplitude, you need to consider the maximum displacement of the brick from the piston, not just the maximum acceleration.

Secondly, it seems like you may have mixed up the values for the period (T) and the angular frequency (w). The angular frequency (w) is equal to 2pi/T, so if the period is 1.19 seconds, the angular frequency should be approximately 5.29 rad/s.

Lastly, when using the equation a(t)=-w^2Acos(wt+phi), the phase angle (phi) should be in radians, not degrees. So, if you are using a value of pi for the phase angle, make sure it is in radians and not degrees.

With these adjustments, you should be able to solve for the amplitude of the brick on the piston. I hope this helps! Good luck with your problem.