Simple harmonic movement problem

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The discussion revolves around a simple harmonic motion problem involving a mass on a vertically oscillating piston with a period of 1 second. The first question addresses the amplitude at which the mass detaches from the piston, with the correct answer being 25 cm, derived from the equation A = g/(4π²). The second question concerns the maximum frequency (2.24 Hz) that allows the mass to remain on the piston, determined by the condition that the piston’s acceleration must exceed gravitational acceleration. Participants emphasize the importance of consistent units and provide guidance on deriving acceleration equations from simple harmonic motion principles. The conversation highlights the relationship between maximum acceleration, amplitude, and gravitational force in this context.
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I would really need help on these questions.

A mass is on a piston that oscillates vertically describing a S.H.M. T= 1s.

1-At which amplitude does the mass comes off the piston?
The answer is supposed to be 25 cm but I don't know how to find the position of the mass when it comes off the piston. I found w but I couldn't write the equation.

2- If the amplitude was 5 cm, what would be the maximum frequency so that the mass remains on the piston?
The answer is f=2,24 Hz
 
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Hint
The mass will leave the piston when the normal reaction between will be zero. This will happen when the acceleration of the piston is more then acceleration due to gravity.
 
Listen to mukundpa he put you on the right track but I want to add a couple of things.

1) Make sure your units match. If you use gravity in m/s^2 your distances must be in meters. (otherwise you can convert gravity to cm/s^2) same thing.

2) You need an eqn. for acceleration
b) if you do not have an eqn for accelration here is how to get one.
(I did this whole thing by comparing shm to circular motion)
Assume that the greatest acceleration is when the piston starts to go down (this drops out any sin from your eqn). Now use centripital acceleration and substitute for velocity (period and frequency will take its place)

3) No need to use w (omega), but it will help the 2Pi disappear.
 
The answer to b) seems to be 2.24Hz and not 2,24 Hz as you have written.
I got this from the equation Aw^2=g
 
Thanks for the help but now I have another question. This is probably stupid but I can't figure out why I get exactly the double for question a. I get 0.49m and the pulsation is equal to 2pi.
 
Acceleration is maximum when the displacement is maximum(amplitude). For minimum amplitude for which the mass just leaves the piston the max. acceleration is g at the max. displacenrnt. Thus

w^2A = g
A = g/4Pi^2 = 9.8/4*9.87 = 0.248m which is close to 25 cm

some time g in m/s/s and pi^2 both are approximated to 10
 
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